∃G ∈ F (G ↔ (G ⊬ F))

There exists a G such that G is logically equivalent to its own
unprovability in F

*If we assume that there is such a G in F that means that*
G is true means there is no sequence of inference steps that satisfies G
in F.
G is false means there is a sequence of inference steps that satisfies G
in F.

*Thus the above G simply does not exist in F*

--
Copyright 2023 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

On 4/21/23 8:33 PM, olcott wrote:
> ∃G ∈ F (G ↔ (G ⊬ F))
>
> There exists a G such that G is logically equivalent to its own
> unprovability in F
>
> *If we assume that there is such a G in F that means that*
> G is true means there is no sequence of inference steps that satisfies G
> in F.
> G is false means there is a sequence of inference steps that satisfies G
> in F.
>
> *Thus the above G simply does not exist in F*
>
>

So?

Since Godel's G isn't of that form, but only can be used to derive a
statment IN META-F that says that G is not provable in F, your argument
says nothing about Godel's G.

Also, you don't understand what those terms mean, because G being true
doesn't mean there is no sequence of inference steps that satisfies G in
F, but there is no FINITE sequence of inference steps that satisfies G in F.

Provable requires a FINITE sequence of steps

Being True just requires a sequence of connective steps, which can be
infinite.

THus, a statement that is True but Unprovable says that there exist a
sequence of connective steps, but that sequence is infinite in length,
so not usable as a proof.

You are just showing that you don't understand the meaning of the terms
you are using.

On 4/21/2023 7:49 PM, Richard Damon wrote:
> On 4/21/23 8:33 PM, olcott wrote:
>> ∃G ∈ F (G ↔ (G ⊬ F))
>>
>> There exists a G such that G is logically equivalent to its own
>> unprovability in F
>>
>> *If we assume that there is such a G in F that means that*
>> G is true means there is no sequence of inference steps that satisfies
>> G in F.
>> G is false means there is a sequence of inference steps that satisfies
>> G in F.
>>
>> *Thus the above G simply does not exist in F*
>>
>>
>
>
> So?
>

I finally learned enough model theory to correctly link provability to
truth in the conventional model theory way.

I finally approximated {G asserts its own unprovability in F}
using conventional math symbols in their conventional way.

> Since Godel's G isn't of that form, but only can be used to derive a
> statment IN META-F that says that G is not provable in F, your argument
> says nothing about Godel's G.
>

F ⊢ GF ↔ ¬ProvF (┌GF┐).
https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
I have finally created a G that is equivalent to
Panu Raatikainen's SEP article.

>
> Also, you don't understand what those terms mean, because G being true
> doesn't mean there is no sequence of inference steps that satisfies G in
> F, but there is no FINITE sequence of inference steps that satisfies G
> in F.
>

∃G ∈ F (G ↔ (G ⊬ F))

Because we can see that every finite or infinite sequence in F that
satisfies the RHS of ↔ contradicts the LHS a powerful F can infer that G
is utterly unsatisfiable even for infinite sequences in this more
powerful F.

> Provable requires a FINITE sequence of steps
>
> Being True just requires a sequence of connective steps, which can be
> infinite.
>
> THus, a statement that is True but Unprovable says that there exist a
> sequence of connective steps, but that sequence is infinite in length,
> so not usable as a proof.
>
>
> You are just showing that you don't understand the meaning of the terms
> you are using.

--
Copyright 2023 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

On 4/21/23 9:41 PM, olcott wrote:
> On 4/21/2023 7:49 PM, Richard Damon wrote:
>> On 4/21/23 8:33 PM, olcott wrote:
>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>
>>> There exists a G such that G is logically equivalent to its own
>>> unprovability in F
>>>
>>> *If we assume that there is such a G in F that means that*
>>> G is true means there is no sequence of inference steps that
>>> satisfies G in F.
>>> G is false means there is a sequence of inference steps that
>>> satisfies G in F.
>>>
>>> *Thus the above G simply does not exist in F*
>>>
>>>
>>
>>
>> So?
>>
>
> I finally learned enough model theory to correctly link provability to
> truth in the conventional model theory way.

Doesn't seem so, you don't seem to understand the difference. You seem
to confuse Truth with Knowledge.

>
> I finally approximated {G asserts its own unprovability in F}
> using conventional math symbols in their conventional way.

Except that isn't what G is, you only think that because you can't
actually understand even the outline of Godel's proof, so you take
pieces out of context.

G never asserts its own unprovability.

The statement that we now have a statement that asserts its own
unprovablity, as a simplification describing a statment DERIVED from G,
and that derivation happens in Meta-F, and is about what can be proven in F.

>
>> Since Godel's G isn't of that form, but only can be used to derive a
>> statment IN META-F that says that G is not provable in F, your
>> argument says nothing about Godel's G.
>>
>
> F ⊢ GF ↔ ¬ProvF (┌GF┐).
> https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
> I have finally created a G that is equivalent to
> Panu Raatikainen's SEP article.

So?

Did you read that article?

>
>>
>> Also, you don't understand what those terms mean, because G being true
>> doesn't mean there is no sequence of inference steps that satisfies G
>> in F, but there is no FINITE sequence of inference steps that
>> satisfies G in F.
>>
>
> ∃G ∈ F (G ↔ (G ⊬ F))
>
> Because we can see that every finite or infinite sequence in F that
> satisfies the RHS of ↔ contradicts the LHS a powerful F can infer that G
> is utterly unsatisfiable even for infinite sequences in this more
> powerful F.

Nope. Show the PROOF.

You don't know HOW to do a proof, you can only do arguement.

You are just too stupid to know how to actually prove a statement.

>
>> Provable requires a FINITE sequence of steps
>>
>> Being True just requires a sequence of connective steps, which can be
>> infinite.
>>
>> THus, a statement that is True but Unprovable says that there exist a
>> sequence of connective steps, but that sequence is infinite in length,
>> so not usable as a proof.
>>
>>
>> You are just showing that you don't understand the meaning of the
>> terms you are using.
>

On 4/21/2023 9:45 PM, Richard Damon wrote:
> On 4/21/23 9:41 PM, olcott wrote:
>> On 4/21/2023 7:49 PM, Richard Damon wrote:
>>> On 4/21/23 8:33 PM, olcott wrote:
>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>
>>>> There exists a G such that G is logically equivalent to its own
>>>> unprovability in F
>>>>
>>>> *If we assume that there is such a G in F that means that*
>>>> G is true means there is no sequence of inference steps that
>>>> satisfies G in F.
>>>> G is false means there is a sequence of inference steps that
>>>> satisfies G in F.
>>>>
>>>> *Thus the above G simply does not exist in F*
>>>>
>>>>
>>>
>>>
>>> So?
>>>
>>
>> I finally learned enough model theory to correctly link provability to
>> truth in the conventional model theory way.
>
> Doesn't seem so, you don't seem to understand the difference. You seem
> to confuse Truth with Knowledge.
>
>>
>> I finally approximated {G asserts its own unprovability in F}
>> using conventional math symbols in their conventional way.
>
> Except that isn't what G is, you only think that because you can't
> actually understand even the outline of Godel's proof, so you take
> pieces out of context.
>
> G never asserts its own unprovability.
>
> The statement that we now have a statement that asserts its own
> unprovablity, as a simplification describing a statment DERIVED from G,
> and that derivation happens in Meta-F, and is about what can be proven
> in F.
>
>>
>>> Since Godel's G isn't of that form, but only can be used to derive a
>>> statment IN META-F that says that G is not provable in F, your
>>> argument says nothing about Godel's G.
>>>
>>
>> F ⊢ GF ↔ ¬ProvF (┌GF┐).
>> https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
>> I have finally created a G that is equivalent to
>> Panu Raatikainen's SEP article.
>
> So?
>
> Did you read that article?
>
>>
>>>
>>> Also, you don't understand what those terms mean, because G being
>>> true doesn't mean there is no sequence of inference steps that
>>> satisfies G in F, but there is no FINITE sequence of inference steps
>>> that satisfies G in F.
>>>
>>
>> ∃G ∈ F (G ↔ (G ⊬ F))
>>
>> Because we can see that every finite or infinite sequence in F that
>> satisfies the RHS of ↔ contradicts the LHS a powerful F can infer that G
>> is utterly unsatisfiable even for infinite sequences in this more
>> powerful F.
>
> Nope. Show the PROOF.
>
> You don't know HOW to do a proof, you can only do arguement.
>

∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically equivalent to its own
unprovability in F

A proof is any sequence of steps that shows that its conclusion is a
necessary consequence of its premises.

∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically equivalent to its own
unprovability in F

If G is true then there is no sequence of inference steps that satisfies
G in F making G untrue.
If G is false then there is a sequence of inference steps that satisfies
G in F making G true.
Because the RHS of ↔ contradicts the LHS there is no such G in F.
Thus the above G simply does not exist in F.

--
Copyright 2023 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

On 4/21/23 11:40 PM, olcott wrote:
> On 4/21/2023 9:45 PM, Richard Damon wrote:
>> On 4/21/23 9:41 PM, olcott wrote:
>>> On 4/21/2023 7:49 PM, Richard Damon wrote:
>>>> On 4/21/23 8:33 PM, olcott wrote:
>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>
>>>>> There exists a G such that G is logically equivalent to its own
>>>>> unprovability in F
>>>>>
>>>>> *If we assume that there is such a G in F that means that*
>>>>> G is true means there is no sequence of inference steps that
>>>>> satisfies G in F.
>>>>> G is false means there is a sequence of inference steps that
>>>>> satisfies G in F.
>>>>>
>>>>> *Thus the above G simply does not exist in F*
>>>>>
>>>>>
>>>>
>>>>
>>>> So?
>>>>
>>>
>>> I finally learned enough model theory to correctly link provability to
>>> truth in the conventional model theory way.
>>
>> Doesn't seem so, you don't seem to understand the difference. You seem
>> to confuse Truth with Knowledge.
>>
>>>
>>> I finally approximated {G asserts its own unprovability in F}
>>> using conventional math symbols in their conventional way.
>>
>> Except that isn't what G is, you only think that because you can't
>> actually understand even the outline of Godel's proof, so you take
>> pieces out of context.
>>
>> G never asserts its own unprovability.
>>
>> The statement that we now have a statement that asserts its own
>> unprovablity, as a simplification describing a statment DERIVED from
>> G, and that derivation happens in Meta-F, and is about what can be
>> proven in F.
>>
>>>
>>>> Since Godel's G isn't of that form, but only can be used to derive a
>>>> statment IN META-F that says that G is not provable in F, your
>>>> argument says nothing about Godel's G.
>>>>
>>>
>>> F ⊢ GF ↔ ¬ProvF (┌GF┐).
>>> https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
>>> I have finally created a G that is equivalent to
>>> Panu Raatikainen's SEP article.
>>
>> So?
>>
>> Did you read that article?
>>
>>>
>>>>
>>>> Also, you don't understand what those terms mean, because G being
>>>> true doesn't mean there is no sequence of inference steps that
>>>> satisfies G in F, but there is no FINITE sequence of inference steps
>>>> that satisfies G in F.
>>>>
>>>
>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>
>>> Because we can see that every finite or infinite sequence in F that
>>> satisfies the RHS of ↔ contradicts the LHS a powerful F can infer that G
>>> is utterly unsatisfiable even for infinite sequences in this more
>>> powerful F.
>>
>> Nope. Show the PROOF.
>>
>> You don't know HOW to do a proof, you can only do arguement.
>>
>
> ∃G ∈ F (G ↔ (G ⊬ F))
> There exists a G in F such that G is logically equivalent to its own
> unprovability in F
>
> A proof is any sequence of steps that shows that its conclusion is a
> necessary consequence of its premises.\

Boy are you wrong.

A proof is a FINITE sequence of steps that shows that a given statement
is a necessary consequence of the defined system.

"Proof" doesn't have a "Premise", it has a system.

The statement may have conditions in it restricting when

>
> ∃G ∈ F (G ↔ (G ⊬ F))
> There exists a G in F such that G is logically equivalent to its own
> unprovability in F
>
> If G is true then there is no sequence of inference steps that satisfies
> G in F making G untrue.

no FINITE sequence, making G UNPROVABLE, and there IS an INFINITE
sequence making it TRUE.

This is possible.

> If G is false then there is a sequence of inference steps that satisfies
> G in F making G true.

If G is false, then there is a finite sequence proving G, which forces G
to be true, thus this is a contradiction.

> Because the RHS of ↔ contradicts the LHS there is no such G in F.
> Thus the above G simply does not exist in F.
>

Nope, because we can have an infinite sequence that isn't finite, G can
be True but not Provable.

You are just showing your stupidity. It goes beyond ignorance, because
it has been explained to you many time, but you show an inability to
learn the simple facts.

You just refuse to use the actual definitions of the system, because you
are just a pathological liar.

On 4/22/2023 6:17 AM, Richard Damon wrote:
> On 4/21/23 11:40 PM, olcott wrote:
>> On 4/21/2023 9:45 PM, Richard Damon wrote:
>>> On 4/21/23 9:41 PM, olcott wrote:
>>>> On 4/21/2023 7:49 PM, Richard Damon wrote:
>>>>> On 4/21/23 8:33 PM, olcott wrote:
>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>
>>>>>> There exists a G such that G is logically equivalent to its own
>>>>>> unprovability in F
>>>>>>
>>>>>> *If we assume that there is such a G in F that means that*
>>>>>> G is true means there is no sequence of inference steps that
>>>>>> satisfies G in F.
>>>>>> G is false means there is a sequence of inference steps that
>>>>>> satisfies G in F.
>>>>>>
>>>>>> *Thus the above G simply does not exist in F*
>>>>>>
>>>>>>
>>>>>
>>>>>
>>>>> So?
>>>>>
>>>>
>>>> I finally learned enough model theory to correctly link provability to
>>>> truth in the conventional model theory way.
>>>
>>> Doesn't seem so, you don't seem to understand the difference. You
>>> seem to confuse Truth with Knowledge.
>>>
>>>>
>>>> I finally approximated {G asserts its own unprovability in F}
>>>> using conventional math symbols in their conventional way.
>>>
>>> Except that isn't what G is, you only think that because you can't
>>> actually understand even the outline of Godel's proof, so you take
>>> pieces out of context.
>>>
>>> G never asserts its own unprovability.
>>>
>>> The statement that we now have a statement that asserts its own
>>> unprovablity, as a simplification describing a statment DERIVED from
>>> G, and that derivation happens in Meta-F, and is about what can be
>>> proven in F.
>>>
>>>>
>>>>> Since Godel's G isn't of that form, but only can be used to derive
>>>>> a statment IN META-F that says that G is not provable in F, your
>>>>> argument says nothing about Godel's G.
>>>>>
>>>>
>>>> F ⊢ GF ↔ ¬ProvF (┌GF┐).
>>>> https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
>>>> I have finally created a G that is equivalent to
>>>> Panu Raatikainen's SEP article.
>>>
>>> So?
>>>
>>> Did you read that article?
>>>
>>>>
>>>>>
>>>>> Also, you don't understand what those terms mean, because G being
>>>>> true doesn't mean there is no sequence of inference steps that
>>>>> satisfies G in F, but there is no FINITE sequence of inference
>>>>> steps that satisfies G in F.
>>>>>
>>>>
>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>
>>>> Because we can see that every finite or infinite sequence in F that
>>>> satisfies the RHS of ↔ contradicts the LHS a powerful F can infer
>>>> that G
>>>> is utterly unsatisfiable even for infinite sequences in this more
>>>> powerful F.
>>>
>>> Nope. Show the PROOF.
>>>
>>> You don't know HOW to do a proof, you can only do arguement.
>>>
>>
>> ∃G ∈ F (G ↔ (G ⊬ F))
>> There exists a G in F such that G is logically equivalent to its own
>> unprovability in F
>>
>> A proof is any sequence of steps that shows that its conclusion is a
>> necessary consequence of its premises.\
>
> Boy are you wrong.
>
> A proof is a FINITE sequence of steps that shows that a given statement
> is a necessary consequence of the defined system.
>
> "Proof" doesn't have a  "Premise", it has a system.
>
> The statement may have conditions in it restricting when
>
>>
>> ∃G ∈ F (G ↔ (G ⊬ F))
>> There exists a G in F such that G is logically equivalent to its own
>> unprovability in F
>>
>> If G is true then there is no sequence of inference steps that
>> satisfies G in F making G untrue.
>
> no FINITE sequence, making G UNPROVABLE, and there IS an INFINITE
> sequence making it TRUE.
>
> This is possible.
>
>> If G is false then there is a sequence of inference steps that
>> satisfies G in F making G true.
>
> If G is false, then there is a finite sequence proving G, which forces G
> to be true, thus this is a contradiction.
>
>> Because the RHS of ↔ contradicts the LHS there is no such G in F.
>> Thus the above G simply does not exist in F.
>>
>
> Nope, because we can have an infinite sequence that isn't finite, G can
> be True but not Provable.
>

If G is false and ↔ is true this makes the RHS false which negates the
RHS making it say (G ⊢ F) which makes G true in F.

--
Copyright 2023 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

On 4/22/23 10:28 AM, olcott wrote:
> On 4/22/2023 6:17 AM, Richard Damon wrote:
>> On 4/21/23 11:40 PM, olcott wrote:
>>> On 4/21/2023 9:45 PM, Richard Damon wrote:
>>>> On 4/21/23 9:41 PM, olcott wrote:
>>>>> On 4/21/2023 7:49 PM, Richard Damon wrote:
>>>>>> On 4/21/23 8:33 PM, olcott wrote:
>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>
>>>>>>> There exists a G such that G is logically equivalent to its own
>>>>>>> unprovability in F
>>>>>>>
>>>>>>> *If we assume that there is such a G in F that means that*
>>>>>>> G is true means there is no sequence of inference steps that
>>>>>>> satisfies G in F.
>>>>>>> G is false means there is a sequence of inference steps that
>>>>>>> satisfies G in F.
>>>>>>>
>>>>>>> *Thus the above G simply does not exist in F*
>>>>>>>
>>>>>>>
>>>>>>
>>>>>>
>>>>>> So?
>>>>>>
>>>>>
>>>>> I finally learned enough model theory to correctly link provability to
>>>>> truth in the conventional model theory way.
>>>>
>>>> Doesn't seem so, you don't seem to understand the difference. You
>>>> seem to confuse Truth with Knowledge.
>>>>
>>>>>
>>>>> I finally approximated {G asserts its own unprovability in F}
>>>>> using conventional math symbols in their conventional way.
>>>>
>>>> Except that isn't what G is, you only think that because you can't
>>>> actually understand even the outline of Godel's proof, so you take
>>>> pieces out of context.
>>>>
>>>> G never asserts its own unprovability.
>>>>
>>>> The statement that we now have a statement that asserts its own
>>>> unprovablity, as a simplification describing a statment DERIVED from
>>>> G, and that derivation happens in Meta-F, and is about what can be
>>>> proven in F.
>>>>
>>>>>
>>>>>> Since Godel's G isn't of that form, but only can be used to derive
>>>>>> a statment IN META-F that says that G is not provable in F, your
>>>>>> argument says nothing about Godel's G.
>>>>>>
>>>>>
>>>>> F ⊢ GF ↔ ¬ProvF (┌GF┐).
>>>>> https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
>>>>> I have finally created a G that is equivalent to
>>>>> Panu Raatikainen's SEP article.
>>>>
>>>> So?
>>>>
>>>> Did you read that article?
>>>>
>>>>>
>>>>>>
>>>>>> Also, you don't understand what those terms mean, because G being
>>>>>> true doesn't mean there is no sequence of inference steps that
>>>>>> satisfies G in F, but there is no FINITE sequence of inference
>>>>>> steps that satisfies G in F.
>>>>>>
>>>>>
>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>
>>>>> Because we can see that every finite or infinite sequence in F that
>>>>> satisfies the RHS of ↔ contradicts the LHS a powerful F can infer
>>>>> that G
>>>>> is utterly unsatisfiable even for infinite sequences in this more
>>>>> powerful F.
>>>>
>>>> Nope. Show the PROOF.
>>>>
>>>> You don't know HOW to do a proof, you can only do arguement.
>>>>
>>>
>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>> There exists a G in F such that G is logically equivalent to its own
>>> unprovability in F
>>>
>>> A proof is any sequence of steps that shows that its conclusion is a
>>> necessary consequence of its premises.\
>>
>> Boy are you wrong.
>>
>> A proof is a FINITE sequence of steps that shows that a given
>> statement is a necessary consequence of the defined system.
>>
>> "Proof" doesn't have a  "Premise", it has a system.
>>
>> The statement may have conditions in it restricting when
>>
>>>
>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>> There exists a G in F such that G is logically equivalent to its own
>>> unprovability in F
>>>
>>> If G is true then there is no sequence of inference steps that
>>> satisfies G in F making G untrue.
>>
>> no FINITE sequence, making G UNPROVABLE, and there IS an INFINITE
>> sequence making it TRUE.
>>
>> This is possible.
>>
>>> If G is false then there is a sequence of inference steps that
>>> satisfies G in F making G true.
>>
>> If G is false, then there is a finite sequence proving G, which forces
>> G to be true, thus this is a contradiction.
>>
>>> Because the RHS of ↔ contradicts the LHS there is no such G in F.
>>> Thus the above G simply does not exist in F.
>>>
>>
>> Nope, because we can have an infinite sequence that isn't finite, G
>> can be True but not Provable.
>>
>
> If G is false and ↔ is true this makes the RHS false which negates the
> RHS making it say (G ⊢ F) which makes G true in F.
>
>

Right, G can't be false, but it can be True.

If G is true, because an INFINITE sequence of steps establishes it, then
it is UNPROVABLE, because no FINITE sequence of steps establishes it.

Thus, it is NOT a contradiction.

You just don't understand what that terms means it seems.

On 4/22/2023 9:38 AM, Richard Damon wrote:
> On 4/22/23 10:28 AM, olcott wrote:
>> On 4/22/2023 6:17 AM, Richard Damon wrote:
>>> On 4/21/23 11:40 PM, olcott wrote:
>>>> On 4/21/2023 9:45 PM, Richard Damon wrote:
>>>>> On 4/21/23 9:41 PM, olcott wrote:
>>>>>> On 4/21/2023 7:49 PM, Richard Damon wrote:
>>>>>>> On 4/21/23 8:33 PM, olcott wrote:
>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>
>>>>>>>> There exists a G such that G is logically equivalent to its own
>>>>>>>> unprovability in F
>>>>>>>>
>>>>>>>> *If we assume that there is such a G in F that means that*
>>>>>>>> G is true means there is no sequence of inference steps that
>>>>>>>> satisfies G in F.
>>>>>>>> G is false means there is a sequence of inference steps that
>>>>>>>> satisfies G in F.
>>>>>>>>
>>>>>>>> *Thus the above G simply does not exist in F*
>>>>>>>>
>>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>> So?
>>>>>>>
>>>>>>
>>>>>> I finally learned enough model theory to correctly link
>>>>>> provability to
>>>>>> truth in the conventional model theory way.
>>>>>
>>>>> Doesn't seem so, you don't seem to understand the difference. You
>>>>> seem to confuse Truth with Knowledge.
>>>>>
>>>>>>
>>>>>> I finally approximated {G asserts its own unprovability in F}
>>>>>> using conventional math symbols in their conventional way.
>>>>>
>>>>> Except that isn't what G is, you only think that because you can't
>>>>> actually understand even the outline of Godel's proof, so you take
>>>>> pieces out of context.
>>>>>
>>>>> G never asserts its own unprovability.
>>>>>
>>>>> The statement that we now have a statement that asserts its own
>>>>> unprovablity, as a simplification describing a statment DERIVED
>>>>> from G, and that derivation happens in Meta-F, and is about what
>>>>> can be proven in F.
>>>>>
>>>>>>
>>>>>>> Since Godel's G isn't of that form, but only can be used to
>>>>>>> derive a statment IN META-F that says that G is not provable in
>>>>>>> F, your argument says nothing about Godel's G.
>>>>>>>
>>>>>>
>>>>>> F ⊢ GF ↔ ¬ProvF (┌GF┐).
>>>>>> https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
>>>>>> I have finally created a G that is equivalent to
>>>>>> Panu Raatikainen's SEP article.
>>>>>
>>>>> So?
>>>>>
>>>>> Did you read that article?
>>>>>
>>>>>>
>>>>>>>
>>>>>>> Also, you don't understand what those terms mean, because G being
>>>>>>> true doesn't mean there is no sequence of inference steps that
>>>>>>> satisfies G in F, but there is no FINITE sequence of inference
>>>>>>> steps that satisfies G in F.
>>>>>>>
>>>>>>
>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>
>>>>>> Because we can see that every finite or infinite sequence in F that
>>>>>> satisfies the RHS of ↔ contradicts the LHS a powerful F can infer
>>>>>> that G
>>>>>> is utterly unsatisfiable even for infinite sequences in this more
>>>>>> powerful F.
>>>>>
>>>>> Nope. Show the PROOF.
>>>>>
>>>>> You don't know HOW to do a proof, you can only do arguement.
>>>>>
>>>>
>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>> There exists a G in F such that G is logically equivalent to its own
>>>> unprovability in F
>>>>
>>>> A proof is any sequence of steps that shows that its conclusion is a
>>>> necessary consequence of its premises.\
>>>
>>> Boy are you wrong.
>>>
>>> A proof is a FINITE sequence of steps that shows that a given
>>> statement is a necessary consequence of the defined system.
>>>
>>> "Proof" doesn't have a  "Premise", it has a system.
>>>
>>> The statement may have conditions in it restricting when
>>>
>>>>
>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>> There exists a G in F such that G is logically equivalent to its own
>>>> unprovability in F
>>>>
>>>> If G is true then there is no sequence of inference steps that
>>>> satisfies G in F making G untrue.
>>>
>>> no FINITE sequence, making G UNPROVABLE, and there IS an INFINITE
>>> sequence making it TRUE.
>>>
>>> This is possible.
>>>
>>>> If G is false then there is a sequence of inference steps that
>>>> satisfies G in F making G true.
>>>
>>> If G is false, then there is a finite sequence proving G, which
>>> forces G to be true, thus this is a contradiction.
>>>
>>>> Because the RHS of ↔ contradicts the LHS there is no such G in F.
>>>> Thus the above G simply does not exist in F.
>>>>
>>>
>>> Nope, because we can have an infinite sequence that isn't finite, G
>>> can be True but not Provable.
>>>
>>
>> If G is false and ↔ is true this makes the RHS false which negates the
>> RHS making it say (G ⊢ F) which makes G true in F.
>>
>>
>
>
> Right, G can't be false, but it can be True.
>

Thus ↔ cannot be satisfied thus no such G exists in F.

--
Copyright 2023 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

On 4/22/2023 6:17 AM, Richard Damon wrote:
> On 4/21/23 11:40 PM, olcott wrote:
>> On 4/21/2023 9:45 PM, Richard Damon wrote:
>>> On 4/21/23 9:41 PM, olcott wrote:
>>>> On 4/21/2023 7:49 PM, Richard Damon wrote:
>>>>> On 4/21/23 8:33 PM, olcott wrote:
>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>
>>>>>> There exists a G such that G is logically equivalent to its own
>>>>>> unprovability in F
>>>>>>
>>>>>> *If we assume that there is such a G in F that means that*
>>>>>> G is true means there is no sequence of inference steps that
>>>>>> satisfies G in F.
>>>>>> G is false means there is a sequence of inference steps that
>>>>>> satisfies G in F.
>>>>>>
>>>>>> *Thus the above G simply does not exist in F*
>>>>>>
>>>>>>
>>>>>
>>>>>
>>>>> So?
>>>>>
>>>>
>>>> I finally learned enough model theory to correctly link provability to
>>>> truth in the conventional model theory way.
>>>
>>> Doesn't seem so, you don't seem to understand the difference. You
>>> seem to confuse Truth with Knowledge.
>>>
>>>>
>>>> I finally approximated {G asserts its own unprovability in F}
>>>> using conventional math symbols in their conventional way.
>>>
>>> Except that isn't what G is, you only think that because you can't
>>> actually understand even the outline of Godel's proof, so you take
>>> pieces out of context.
>>>
>>> G never asserts its own unprovability.
>>>
>>> The statement that we now have a statement that asserts its own
>>> unprovablity, as a simplification describing a statment DERIVED from
>>> G, and that derivation happens in Meta-F, and is about what can be
>>> proven in F.
>>>
>>>>
>>>>> Since Godel's G isn't of that form, but only can be used to derive
>>>>> a statment IN META-F that says that G is not provable in F, your
>>>>> argument says nothing about Godel's G.
>>>>>
>>>>
>>>> F ⊢ GF ↔ ¬ProvF (┌GF┐).
>>>> https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
>>>> I have finally created a G that is equivalent to
>>>> Panu Raatikainen's SEP article.
>>>
>>> So?
>>>
>>> Did you read that article?
>>>
>>>>
>>>>>
>>>>> Also, you don't understand what those terms mean, because G being
>>>>> true doesn't mean there is no sequence of inference steps that
>>>>> satisfies G in F, but there is no FINITE sequence of inference
>>>>> steps that satisfies G in F.
>>>>>
>>>>
>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>
>>>> Because we can see that every finite or infinite sequence in F that
>>>> satisfies the RHS of ↔ contradicts the LHS a powerful F can infer
>>>> that G
>>>> is utterly unsatisfiable even for infinite sequences in this more
>>>> powerful F.
>>>
>>> Nope. Show the PROOF.
>>>
>>> You don't know HOW to do a proof, you can only do arguement.
>>>
>>
>> ∃G ∈ F (G ↔ (G ⊬ F))
>> There exists a G in F such that G is logically equivalent to its own
>> unprovability in F
>>
>> A proof is any sequence of steps that shows that its conclusion is a
>> necessary consequence of its premises.\
>
> Boy are you wrong.
>
> A proof is a FINITE sequence of steps that shows that a given statement
> is a necessary consequence of the defined system.
>
> "Proof" doesn't have a  "Premise", it has a system.
>
> The statement may have conditions in it restricting when
>
>>
>> ∃G ∈ F (G ↔ (G ⊬ F))
>> There exists a G in F such that G is logically equivalent to its own
>> unprovability in F
>>
>> If G is true then there is no sequence of inference steps that
>> satisfies G in F making G untrue.
>
> no FINITE sequence, making G UNPROVABLE, and there IS an INFINITE
> sequence making it TRUE.
>
> This is possible.
>
>> If G is false then there is a sequence of inference steps that
>> satisfies G in F making G true.
>
> If G is false, then there is a finite sequence proving G, which forces G
> to be true, thus this is a contradiction.
>

When you agree with me on this very important point how does it make
sense to denigrate me further down?

Proving that there is at least one case where ↔ is not satisfied proves
that there is no such G in F that satisfies ↔ in both cases, thus no
such G exists in F. Thus Gödel’s "incompleteness" theorem is transformed
into Olcott's can't prove a contradiction theorem.

*It is like you are saying that I am a liar because I tell the truth*
I am never a Liar because Truth is the most important thing in my life
much more important than love.

Truth is the anchor of the basis of mathematically optimizing existence
thus creating paradise on Earth for all.

--
Copyright 2023 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

On 4/22/23 10:46 AM, olcott wrote:
> On 4/22/2023 6:17 AM, Richard Damon wrote:
>> On 4/21/23 11:40 PM, olcott wrote:
>>> On 4/21/2023 9:45 PM, Richard Damon wrote:
>>>> On 4/21/23 9:41 PM, olcott wrote:
>>>>> On 4/21/2023 7:49 PM, Richard Damon wrote:
>>>>>> On 4/21/23 8:33 PM, olcott wrote:
>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>
>>>>>>> There exists a G such that G is logically equivalent to its own
>>>>>>> unprovability in F
>>>>>>>
>>>>>>> *If we assume that there is such a G in F that means that*
>>>>>>> G is true means there is no sequence of inference steps that
>>>>>>> satisfies G in F.
>>>>>>> G is false means there is a sequence of inference steps that
>>>>>>> satisfies G in F.
>>>>>>>
>>>>>>> *Thus the above G simply does not exist in F*
>>>>>>>
>>>>>>>
>>>>>>
>>>>>>
>>>>>> So?
>>>>>>
>>>>>
>>>>> I finally learned enough model theory to correctly link provability to
>>>>> truth in the conventional model theory way.
>>>>
>>>> Doesn't seem so, you don't seem to understand the difference. You
>>>> seem to confuse Truth with Knowledge.
>>>>
>>>>>
>>>>> I finally approximated {G asserts its own unprovability in F}
>>>>> using conventional math symbols in their conventional way.
>>>>
>>>> Except that isn't what G is, you only think that because you can't
>>>> actually understand even the outline of Godel's proof, so you take
>>>> pieces out of context.
>>>>
>>>> G never asserts its own unprovability.
>>>>
>>>> The statement that we now have a statement that asserts its own
>>>> unprovablity, as a simplification describing a statment DERIVED from
>>>> G, and that derivation happens in Meta-F, and is about what can be
>>>> proven in F.
>>>>
>>>>>
>>>>>> Since Godel's G isn't of that form, but only can be used to derive
>>>>>> a statment IN META-F that says that G is not provable in F, your
>>>>>> argument says nothing about Godel's G.
>>>>>>
>>>>>
>>>>> F ⊢ GF ↔ ¬ProvF (┌GF┐).
>>>>> https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
>>>>> I have finally created a G that is equivalent to
>>>>> Panu Raatikainen's SEP article.
>>>>
>>>> So?
>>>>
>>>> Did you read that article?
>>>>
>>>>>
>>>>>>
>>>>>> Also, you don't understand what those terms mean, because G being
>>>>>> true doesn't mean there is no sequence of inference steps that
>>>>>> satisfies G in F, but there is no FINITE sequence of inference
>>>>>> steps that satisfies G in F.
>>>>>>
>>>>>
>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>
>>>>> Because we can see that every finite or infinite sequence in F that
>>>>> satisfies the RHS of ↔ contradicts the LHS a powerful F can infer
>>>>> that G
>>>>> is utterly unsatisfiable even for infinite sequences in this more
>>>>> powerful F.
>>>>
>>>> Nope. Show the PROOF.
>>>>
>>>> You don't know HOW to do a proof, you can only do arguement.
>>>>
>>>
>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>> There exists a G in F such that G is logically equivalent to its own
>>> unprovability in F
>>>
>>> A proof is any sequence of steps that shows that its conclusion is a
>>> necessary consequence of its premises.\
>>
>> Boy are you wrong.
>>
>> A proof is a FINITE sequence of steps that shows that a given
>> statement is a necessary consequence of the defined system.
>>
>> "Proof" doesn't have a  "Premise", it has a system.
>>
>> The statement may have conditions in it restricting when
>>
>>>
>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>> There exists a G in F such that G is logically equivalent to its own
>>> unprovability in F
>>>
>>> If G is true then there is no sequence of inference steps that
>>> satisfies G in F making G untrue.
>>
>> no FINITE sequence, making G UNPROVABLE, and there IS an INFINITE
>> sequence making it TRUE.
>>
>> This is possible.
>>
>>> If G is false then there is a sequence of inference steps that
>>> satisfies G in F making G true.
>>
>> If G is false, then there is a finite sequence proving G, which forces
>> G to be true, thus this is a contradiction.
>>
>
> When you agree with me on this very important point how does it make
> sense to denigrate me further down?
>
> Proving that there is at least one case where ↔ is not satisfied proves
> that there is no such G in F that satisfies ↔ in both cases, thus no
> such G exists in F. Thus Gödel’s "incompleteness" theorem is transformed
> into Olcott's can't prove a contradiction theorem.

Except that you haven't proven that the case ACTUAL EXISTS.

Values assigned to variables that are IMPOSSIBLE do not need to make sense.

>
> *It is like you are saying that I am a liar because I tell the truth*
> I am never a Liar because Truth is the most important thing in my life
> much more important than love.

No, you are a liar because you tell lies that is things that are untrue.

They may be true in a fantasy world that doesn't actually exist, but
that doesn't make them actually true here in reality.

>
> Truth is the anchor of the basis of mathematically optimizing existence
> thus creating paradise on Earth for all.
>

Right, and since you think truth can come out of impossible situations,
you don't actually understand what truth is.

On 4/22/23 10:48 AM, olcott wrote:
> On 4/22/2023 9:38 AM, Richard Damon wrote:
>> On 4/22/23 10:28 AM, olcott wrote:
>>> On 4/22/2023 6:17 AM, Richard Damon wrote:
>>>> On 4/21/23 11:40 PM, olcott wrote:
>>>>> On 4/21/2023 9:45 PM, Richard Damon wrote:
>>>>>> On 4/21/23 9:41 PM, olcott wrote:
>>>>>>> On 4/21/2023 7:49 PM, Richard Damon wrote:
>>>>>>>> On 4/21/23 8:33 PM, olcott wrote:
>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>>
>>>>>>>>> There exists a G such that G is logically equivalent to its own
>>>>>>>>> unprovability in F
>>>>>>>>>
>>>>>>>>> *If we assume that there is such a G in F that means that*
>>>>>>>>> G is true means there is no sequence of inference steps that
>>>>>>>>> satisfies G in F.
>>>>>>>>> G is false means there is a sequence of inference steps that
>>>>>>>>> satisfies G in F.
>>>>>>>>>
>>>>>>>>> *Thus the above G simply does not exist in F*
>>>>>>>>>
>>>>>>>>>
>>>>>>>>
>>>>>>>>
>>>>>>>> So?
>>>>>>>>
>>>>>>>
>>>>>>> I finally learned enough model theory to correctly link
>>>>>>> provability to
>>>>>>> truth in the conventional model theory way.
>>>>>>
>>>>>> Doesn't seem so, you don't seem to understand the difference. You
>>>>>> seem to confuse Truth with Knowledge.
>>>>>>
>>>>>>>
>>>>>>> I finally approximated {G asserts its own unprovability in F}
>>>>>>> using conventional math symbols in their conventional way.
>>>>>>
>>>>>> Except that isn't what G is, you only think that because you can't
>>>>>> actually understand even the outline of Godel's proof, so you take
>>>>>> pieces out of context.
>>>>>>
>>>>>> G never asserts its own unprovability.
>>>>>>
>>>>>> The statement that we now have a statement that asserts its own
>>>>>> unprovablity, as a simplification describing a statment DERIVED
>>>>>> from G, and that derivation happens in Meta-F, and is about what
>>>>>> can be proven in F.
>>>>>>
>>>>>>>
>>>>>>>> Since Godel's G isn't of that form, but only can be used to
>>>>>>>> derive a statment IN META-F that says that G is not provable in
>>>>>>>> F, your argument says nothing about Godel's G.
>>>>>>>>
>>>>>>>
>>>>>>> F ⊢ GF ↔ ¬ProvF (┌GF┐).
>>>>>>> https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
>>>>>>> I have finally created a G that is equivalent to
>>>>>>> Panu Raatikainen's SEP article.
>>>>>>
>>>>>> So?
>>>>>>
>>>>>> Did you read that article?
>>>>>>
>>>>>>>
>>>>>>>>
>>>>>>>> Also, you don't understand what those terms mean, because G
>>>>>>>> being true doesn't mean there is no sequence of inference steps
>>>>>>>> that satisfies G in F, but there is no FINITE sequence of
>>>>>>>> inference steps that satisfies G in F.
>>>>>>>>
>>>>>>>
>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>
>>>>>>> Because we can see that every finite or infinite sequence in F that
>>>>>>> satisfies the RHS of ↔ contradicts the LHS a powerful F can infer
>>>>>>> that G
>>>>>>> is utterly unsatisfiable even for infinite sequences in this more
>>>>>>> powerful F.
>>>>>>
>>>>>> Nope. Show the PROOF.
>>>>>>
>>>>>> You don't know HOW to do a proof, you can only do arguement.
>>>>>>
>>>>>
>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>> There exists a G in F such that G is logically equivalent to its
>>>>> own unprovability in F
>>>>>
>>>>> A proof is any sequence of steps that shows that its conclusion is a
>>>>> necessary consequence of its premises.\
>>>>
>>>> Boy are you wrong.
>>>>
>>>> A proof is a FINITE sequence of steps that shows that a given
>>>> statement is a necessary consequence of the defined system.
>>>>
>>>> "Proof" doesn't have a  "Premise", it has a system.
>>>>
>>>> The statement may have conditions in it restricting when
>>>>
>>>>>
>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>> There exists a G in F such that G is logically equivalent to its
>>>>> own unprovability in F
>>>>>
>>>>> If G is true then there is no sequence of inference steps that
>>>>> satisfies G in F making G untrue.
>>>>
>>>> no FINITE sequence, making G UNPROVABLE, and there IS an INFINITE
>>>> sequence making it TRUE.
>>>>
>>>> This is possible.
>>>>
>>>>> If G is false then there is a sequence of inference steps that
>>>>> satisfies G in F making G true.
>>>>
>>>> If G is false, then there is a finite sequence proving G, which
>>>> forces G to be true, thus this is a contradiction.
>>>>
>>>>> Because the RHS of ↔ contradicts the LHS there is no such G in F.
>>>>> Thus the above G simply does not exist in F.
>>>>>
>>>>
>>>> Nope, because we can have an infinite sequence that isn't finite, G
>>>> can be True but not Provable.
>>>>
>>>
>>> If G is false and ↔ is true this makes the RHS false which negates
>>> the RHS making it say (G ⊢ F) which makes G true in F.
>>>
>>>
>>
>>
>> Right, G can't be false, but it can be True.
>>
>
> Thus ↔ cannot be satisfied thus no such G exists in F.
>

Why do you say that?

I don't think you know what you terms mean.

There exists a G in F such that G is true if and only if G is Unprovable.

A G that is ALWAYS True and ALWAYS unprovable statisfies that relationship.

"A if and only if B" doesn't requre that the case that neither A and B
being true exists.

Peter Olcott wrote:
....
> *It is like you are saying that I am a liar because I tell the truth*
> I am never a Liar because Truth is the most important thing in my life
> much more important than love.

If you actually believe in your claims, your technically not a liar, but
a crank. This doesn't change much to the fact that your claims are
wrong, and provably wrong.

Anyway you've used a lot of rhetorical sophistic evasions, this may
suggest that you don't really believe in your fallacious claims.

> Truth is the anchor of the basis of mathematically optimizing existence
> thus creating paradise on Earth for all.

You are demented, Peter. And a criminal too.

On 4/22/2023 9:57 AM, Richard Damon wrote:
> On 4/22/23 10:48 AM, olcott wrote:
>> On 4/22/2023 9:38 AM, Richard Damon wrote:
>>> On 4/22/23 10:28 AM, olcott wrote:
>>>> On 4/22/2023 6:17 AM, Richard Damon wrote:
>>>>> On 4/21/23 11:40 PM, olcott wrote:
>>>>>> On 4/21/2023 9:45 PM, Richard Damon wrote:
>>>>>>> On 4/21/23 9:41 PM, olcott wrote:
>>>>>>>> On 4/21/2023 7:49 PM, Richard Damon wrote:
>>>>>>>>> On 4/21/23 8:33 PM, olcott wrote:
>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>>>
>>>>>>>>>> There exists a G such that G is logically equivalent to its
>>>>>>>>>> own unprovability in F
>>>>>>>>>>
>>>>>>>>>> *If we assume that there is such a G in F that means that*
>>>>>>>>>> G is true means there is no sequence of inference steps that
>>>>>>>>>> satisfies G in F.
>>>>>>>>>> G is false means there is a sequence of inference steps that
>>>>>>>>>> satisfies G in F.
>>>>>>>>>>
>>>>>>>>>> *Thus the above G simply does not exist in F*
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>>
>>>>>>>>> So?
>>>>>>>>>
>>>>>>>>
>>>>>>>> I finally learned enough model theory to correctly link
>>>>>>>> provability to
>>>>>>>> truth in the conventional model theory way.
>>>>>>>
>>>>>>> Doesn't seem so, you don't seem to understand the difference. You
>>>>>>> seem to confuse Truth with Knowledge.
>>>>>>>
>>>>>>>>
>>>>>>>> I finally approximated {G asserts its own unprovability in F}
>>>>>>>> using conventional math symbols in their conventional way.
>>>>>>>
>>>>>>> Except that isn't what G is, you only think that because you
>>>>>>> can't actually understand even the outline of Godel's proof, so
>>>>>>> you take pieces out of context.
>>>>>>>
>>>>>>> G never asserts its own unprovability.
>>>>>>>
>>>>>>> The statement that we now have a statement that asserts its own
>>>>>>> unprovablity, as a simplification describing a statment DERIVED
>>>>>>> from G, and that derivation happens in Meta-F, and is about what
>>>>>>> can be proven in F.
>>>>>>>
>>>>>>>>
>>>>>>>>> Since Godel's G isn't of that form, but only can be used to
>>>>>>>>> derive a statment IN META-F that says that G is not provable in
>>>>>>>>> F, your argument says nothing about Godel's G.
>>>>>>>>>
>>>>>>>>
>>>>>>>> F ⊢ GF ↔ ¬ProvF (┌GF┐).
>>>>>>>> https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
>>>>>>>> I have finally created a G that is equivalent to
>>>>>>>> Panu Raatikainen's SEP article.
>>>>>>>
>>>>>>> So?
>>>>>>>
>>>>>>> Did you read that article?
>>>>>>>
>>>>>>>>
>>>>>>>>>
>>>>>>>>> Also, you don't understand what those terms mean, because G
>>>>>>>>> being true doesn't mean there is no sequence of inference steps
>>>>>>>>> that satisfies G in F, but there is no FINITE sequence of
>>>>>>>>> inference steps that satisfies G in F.
>>>>>>>>>
>>>>>>>>
>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>
>>>>>>>> Because we can see that every finite or infinite sequence in F that
>>>>>>>> satisfies the RHS of ↔ contradicts the LHS a powerful F can
>>>>>>>> infer that G
>>>>>>>> is utterly unsatisfiable even for infinite sequences in this more
>>>>>>>> powerful F.
>>>>>>>
>>>>>>> Nope. Show the PROOF.
>>>>>>>
>>>>>>> You don't know HOW to do a proof, you can only do arguement.
>>>>>>>
>>>>>>
>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>> There exists a G in F such that G is logically equivalent to its
>>>>>> own unprovability in F
>>>>>>
>>>>>> A proof is any sequence of steps that shows that its conclusion is a
>>>>>> necessary consequence of its premises.\
>>>>>
>>>>> Boy are you wrong.
>>>>>
>>>>> A proof is a FINITE sequence of steps that shows that a given
>>>>> statement is a necessary consequence of the defined system.
>>>>>
>>>>> "Proof" doesn't have a  "Premise", it has a system.
>>>>>
>>>>> The statement may have conditions in it restricting when
>>>>>
>>>>>>
>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>> There exists a G in F such that G is logically equivalent to its
>>>>>> own unprovability in F
>>>>>>
>>>>>> If G is true then there is no sequence of inference steps that
>>>>>> satisfies G in F making G untrue.
>>>>>
>>>>> no FINITE sequence, making G UNPROVABLE, and there IS an INFINITE
>>>>> sequence making it TRUE.
>>>>>
>>>>> This is possible.
>>>>>
>>>>>> If G is false then there is a sequence of inference steps that
>>>>>> satisfies G in F making G true.
>>>>>
>>>>> If G is false, then there is a finite sequence proving G, which
>>>>> forces G to be true, thus this is a contradiction.
>>>>>
>>>>>> Because the RHS of ↔ contradicts the LHS there is no such G in F.
>>>>>> Thus the above G simply does not exist in F.
>>>>>>
>>>>>
>>>>> Nope, because we can have an infinite sequence that isn't finite, G
>>>>> can be True but not Provable.
>>>>>
>>>>
>>>> If G is false and ↔ is true this makes the RHS false which negates
>>>> the RHS making it say (G ⊢ F) which makes G true in F.
>>>>
>>>>
>>>
>>>
>>> Right, G can't be false, but it can be True.
>>>
>>
>> Thus ↔ cannot be satisfied thus no such G exists in F.
>>
>
> Why do you say that?
>
> I don't think you know what you terms mean.
>
> There exists a G in F such that G is true if and only if G is Unprovable.
>

Logical equality
p q p ↔ q
T T T // G is true if and only if G is Unprovable.
T F F //
F T F //
F F T // G is false if and only if G is Provable.
https://en.wikipedia.org/wiki/Truth_table#Logical_equality

Row(1) There exists a G in F such that G is true if and only if G is
unprovable in F making G unsatisfied thus untrue in F.

Row(4) There exists a G in F such that G is false if and only if G is
provable in F making G satisfied thus true in F.

If either Row(1) or Row(4) are unsatisfied then ↔ is false.

> A G that is ALWAYS True and ALWAYS unprovable statisfies that relationship.
>
> "A if and only if B" doesn't requre that the case that neither A and B
> being true exists.

--
Copyright 2023 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

On 4/22/23 11:39 AM, olcott wrote:
> On 4/22/2023 9:57 AM, Richard Damon wrote:
>> On 4/22/23 10:48 AM, olcott wrote:
>>> On 4/22/2023 9:38 AM, Richard Damon wrote:
>>>> On 4/22/23 10:28 AM, olcott wrote:
>>>>> On 4/22/2023 6:17 AM, Richard Damon wrote:
>>>>>> On 4/21/23 11:40 PM, olcott wrote:
>>>>>>> On 4/21/2023 9:45 PM, Richard Damon wrote:
>>>>>>>> On 4/21/23 9:41 PM, olcott wrote:
>>>>>>>>> On 4/21/2023 7:49 PM, Richard Damon wrote:
>>>>>>>>>> On 4/21/23 8:33 PM, olcott wrote:
>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>>>>
>>>>>>>>>>> There exists a G such that G is logically equivalent to its
>>>>>>>>>>> own unprovability in F
>>>>>>>>>>>
>>>>>>>>>>> *If we assume that there is such a G in F that means that*
>>>>>>>>>>> G is true means there is no sequence of inference steps that
>>>>>>>>>>> satisfies G in F.
>>>>>>>>>>> G is false means there is a sequence of inference steps that
>>>>>>>>>>> satisfies G in F.
>>>>>>>>>>>
>>>>>>>>>>> *Thus the above G simply does not exist in F*
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> So?
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> I finally learned enough model theory to correctly link
>>>>>>>>> provability to
>>>>>>>>> truth in the conventional model theory way.
>>>>>>>>
>>>>>>>> Doesn't seem so, you don't seem to understand the difference.
>>>>>>>> You seem to confuse Truth with Knowledge.
>>>>>>>>
>>>>>>>>>
>>>>>>>>> I finally approximated {G asserts its own unprovability in F}
>>>>>>>>> using conventional math symbols in their conventional way.
>>>>>>>>
>>>>>>>> Except that isn't what G is, you only think that because you
>>>>>>>> can't actually understand even the outline of Godel's proof, so
>>>>>>>> you take pieces out of context.
>>>>>>>>
>>>>>>>> G never asserts its own unprovability.
>>>>>>>>
>>>>>>>> The statement that we now have a statement that asserts its own
>>>>>>>> unprovablity, as a simplification describing a statment DERIVED
>>>>>>>> from G, and that derivation happens in Meta-F, and is about what
>>>>>>>> can be proven in F.
>>>>>>>>
>>>>>>>>>
>>>>>>>>>> Since Godel's G isn't of that form, but only can be used to
>>>>>>>>>> derive a statment IN META-F that says that G is not provable
>>>>>>>>>> in F, your argument says nothing about Godel's G.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> F ⊢ GF ↔ ¬ProvF (┌GF┐).
>>>>>>>>> https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
>>>>>>>>> I have finally created a G that is equivalent to
>>>>>>>>> Panu Raatikainen's SEP article.
>>>>>>>>
>>>>>>>> So?
>>>>>>>>
>>>>>>>> Did you read that article?
>>>>>>>>
>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> Also, you don't understand what those terms mean, because G
>>>>>>>>>> being true doesn't mean there is no sequence of inference
>>>>>>>>>> steps that satisfies G in F, but there is no FINITE sequence
>>>>>>>>>> of inference steps that satisfies G in F.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>>
>>>>>>>>> Because we can see that every finite or infinite sequence in F
>>>>>>>>> that
>>>>>>>>> satisfies the RHS of ↔ contradicts the LHS a powerful F can
>>>>>>>>> infer that G
>>>>>>>>> is utterly unsatisfiable even for infinite sequences in this more
>>>>>>>>> powerful F.
>>>>>>>>
>>>>>>>> Nope. Show the PROOF.
>>>>>>>>
>>>>>>>> You don't know HOW to do a proof, you can only do arguement.
>>>>>>>>
>>>>>>>
>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>> There exists a G in F such that G is logically equivalent to its
>>>>>>> own unprovability in F
>>>>>>>
>>>>>>> A proof is any sequence of steps that shows that its conclusion is a
>>>>>>> necessary consequence of its premises.\
>>>>>>
>>>>>> Boy are you wrong.
>>>>>>
>>>>>> A proof is a FINITE sequence of steps that shows that a given
>>>>>> statement is a necessary consequence of the defined system.
>>>>>>
>>>>>> "Proof" doesn't have a  "Premise", it has a system.
>>>>>>
>>>>>> The statement may have conditions in it restricting when
>>>>>>
>>>>>>>
>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>> There exists a G in F such that G is logically equivalent to its
>>>>>>> own unprovability in F
>>>>>>>
>>>>>>> If G is true then there is no sequence of inference steps that
>>>>>>> satisfies G in F making G untrue.
>>>>>>
>>>>>> no FINITE sequence, making G UNPROVABLE, and there IS an INFINITE
>>>>>> sequence making it TRUE.
>>>>>>
>>>>>> This is possible.
>>>>>>
>>>>>>> If G is false then there is a sequence of inference steps that
>>>>>>> satisfies G in F making G true.
>>>>>>
>>>>>> If G is false, then there is a finite sequence proving G, which
>>>>>> forces G to be true, thus this is a contradiction.
>>>>>>
>>>>>>> Because the RHS of ↔ contradicts the LHS there is no such G in F.
>>>>>>> Thus the above G simply does not exist in F.
>>>>>>>
>>>>>>
>>>>>> Nope, because we can have an infinite sequence that isn't finite,
>>>>>> G can be True but not Provable.
>>>>>>
>>>>>
>>>>> If G is false and ↔ is true this makes the RHS false which negates
>>>>> the RHS making it say (G ⊢ F) which makes G true in F.
>>>>>
>>>>>
>>>>
>>>>
>>>> Right, G can't be false, but it can be True.
>>>>
>>>
>>> Thus ↔ cannot be satisfied thus no such G exists in F.
>>>
>>
>> Why do you say that?
>>
>> I don't think you know what you terms mean.
>>
>> There exists a G in F such that G is true if and only if G is Unprovable.
>>
>
> Logical equality
> p q p ↔ q
> T T   T // G is true if and only if G is Unprovable.
> T F   F //
> F T   F //
> F F   T // G is false if and only if G is Provable.
> https://en.wikipedia.org/wiki/Truth_table#Logical_equality
>
> Row(1) There exists a G in F such that G is true if and only if G is
> unprovable in F making G unsatisfied thus untrue in F.
>
> Row(4) There exists a G in F such that G is false if and only if G is
> provable in F making G satisfied thus true in F.
>
> If either Row(1) or Row(4) are unsatisfied then ↔ is false.

On 4/22/2023 11:12 AM, Richard Damon wrote:
> On 4/22/23 11:39 AM, olcott wrote:
>> On 4/22/2023 9:57 AM, Richard Damon wrote:
>>> On 4/22/23 10:48 AM, olcott wrote:
>>>> On 4/22/2023 9:38 AM, Richard Damon wrote:
>>>>> On 4/22/23 10:28 AM, olcott wrote:
>>>>>> On 4/22/2023 6:17 AM, Richard Damon wrote:
>>>>>>> On 4/21/23 11:40 PM, olcott wrote:
>>>>>>>> On 4/21/2023 9:45 PM, Richard Damon wrote:
>>>>>>>>> On 4/21/23 9:41 PM, olcott wrote:
>>>>>>>>>> On 4/21/2023 7:49 PM, Richard Damon wrote:
>>>>>>>>>>> On 4/21/23 8:33 PM, olcott wrote:
>>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>>>>>
>>>>>>>>>>>> There exists a G such that G is logically equivalent to its
>>>>>>>>>>>> own unprovability in F
>>>>>>>>>>>>
>>>>>>>>>>>> *If we assume that there is such a G in F that means that*
>>>>>>>>>>>> G is true means there is no sequence of inference steps that
>>>>>>>>>>>> satisfies G in F.
>>>>>>>>>>>> G is false means there is a sequence of inference steps that
>>>>>>>>>>>> satisfies G in F.
>>>>>>>>>>>>
>>>>>>>>>>>> *Thus the above G simply does not exist in F*
>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> So?
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> I finally learned enough model theory to correctly link
>>>>>>>>>> provability to
>>>>>>>>>> truth in the conventional model theory way.
>>>>>>>>>
>>>>>>>>> Doesn't seem so, you don't seem to understand the difference.
>>>>>>>>> You seem to confuse Truth with Knowledge.
>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> I finally approximated {G asserts its own unprovability in F}
>>>>>>>>>> using conventional math symbols in their conventional way.
>>>>>>>>>
>>>>>>>>> Except that isn't what G is, you only think that because you
>>>>>>>>> can't actually understand even the outline of Godel's proof, so
>>>>>>>>> you take pieces out of context.
>>>>>>>>>
>>>>>>>>> G never asserts its own unprovability.
>>>>>>>>>
>>>>>>>>> The statement that we now have a statement that asserts its own
>>>>>>>>> unprovablity, as a simplification describing a statment DERIVED
>>>>>>>>> from G, and that derivation happens in Meta-F, and is about
>>>>>>>>> what can be proven in F.
>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>>> Since Godel's G isn't of that form, but only can be used to
>>>>>>>>>>> derive a statment IN META-F that says that G is not provable
>>>>>>>>>>> in F, your argument says nothing about Godel's G.
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> F ⊢ GF ↔ ¬ProvF (┌GF┐).
>>>>>>>>>> https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
>>>>>>>>>> I have finally created a G that is equivalent to
>>>>>>>>>> Panu Raatikainen's SEP article.
>>>>>>>>>
>>>>>>>>> So?
>>>>>>>>>
>>>>>>>>> Did you read that article?
>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> Also, you don't understand what those terms mean, because G
>>>>>>>>>>> being true doesn't mean there is no sequence of inference
>>>>>>>>>>> steps that satisfies G in F, but there is no FINITE sequence
>>>>>>>>>>> of inference steps that satisfies G in F.
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>>>
>>>>>>>>>> Because we can see that every finite or infinite sequence in F
>>>>>>>>>> that
>>>>>>>>>> satisfies the RHS of ↔ contradicts the LHS a powerful F can
>>>>>>>>>> infer that G
>>>>>>>>>> is utterly unsatisfiable even for infinite sequences in this more
>>>>>>>>>> powerful F.
>>>>>>>>>
>>>>>>>>> Nope. Show the PROOF.
>>>>>>>>>
>>>>>>>>> You don't know HOW to do a proof, you can only do arguement.
>>>>>>>>>
>>>>>>>>
>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>> There exists a G in F such that G is logically equivalent to its
>>>>>>>> own unprovability in F
>>>>>>>>
>>>>>>>> A proof is any sequence of steps that shows that its conclusion
>>>>>>>> is a
>>>>>>>> necessary consequence of its premises.\
>>>>>>>
>>>>>>> Boy are you wrong.
>>>>>>>
>>>>>>> A proof is a FINITE sequence of steps that shows that a given
>>>>>>> statement is a necessary consequence of the defined system.
>>>>>>>
>>>>>>> "Proof" doesn't have a  "Premise", it has a system.
>>>>>>>
>>>>>>> The statement may have conditions in it restricting when
>>>>>>>
>>>>>>>>
>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>> There exists a G in F such that G is logically equivalent to its
>>>>>>>> own unprovability in F
>>>>>>>>
>>>>>>>> If G is true then there is no sequence of inference steps that
>>>>>>>> satisfies G in F making G untrue.
>>>>>>>
>>>>>>> no FINITE sequence, making G UNPROVABLE, and there IS an INFINITE
>>>>>>> sequence making it TRUE.
>>>>>>>
>>>>>>> This is possible.
>>>>>>>
>>>>>>>> If G is false then there is a sequence of inference steps that
>>>>>>>> satisfies G in F making G true.
>>>>>>>
>>>>>>> If G is false, then there is a finite sequence proving G, which
>>>>>>> forces G to be true, thus this is a contradiction.
>>>>>>>
>>>>>>>> Because the RHS of ↔ contradicts the LHS there is no such G in F.
>>>>>>>> Thus the above G simply does not exist in F.
>>>>>>>>
>>>>>>>
>>>>>>> Nope, because we can have an infinite sequence that isn't finite,
>>>>>>> G can be True but not Provable.
>>>>>>>
>>>>>>
>>>>>> If G is false and ↔ is true this makes the RHS false which negates
>>>>>> the RHS making it say (G ⊢ F) which makes G true in F.
>>>>>>
>>>>>>
>>>>>
>>>>>
>>>>> Right, G can't be false, but it can be True.
>>>>>
>>>>
>>>> Thus ↔ cannot be satisfied thus no such G exists in F.
>>>>
>>>
>>> Why do you say that?
>>>
>>> I don't think you know what you terms mean.
>>>
>>> There exists a G in F such that G is true if and only if G is
>>> Unprovable.
>>>
>>
>> Logical equality
>> p q p ↔ q
>> T T   T // G is true if and only if G is Unprovable.
>> T F   F //
>> F T   F //
>> F F   T // G is false if and only if G is Provable.
>> https://en.wikipedia.org/wiki/Truth_table#Logical_equality
>>
>> Row(1) There exists a G in F such that G is true if and only if G is
>> unprovable in F making G unsatisfied thus untrue in F.
>>
>> Row(4) There exists a G in F such that G is false if and only if G is
>> provable in F making G satisfied thus true in F.
>>
>> If either Row(1) or Row(4) are unsatisfied then ↔ is false.
>
> But if neither row values can ACTUALLY EXIST, then the equality is true.
>
If either Row(1) or Row(4) cannot have the same value for p and q
(for whatever reason) then ↔ is unsatisfied and no such G exists in F.

On 4/22/23 12:27 PM, olcott wrote:
> On 4/22/2023 11:12 AM, Richard Damon wrote:
>> On 4/22/23 11:39 AM, olcott wrote:
>>> On 4/22/2023 9:57 AM, Richard Damon wrote:
>>>> On 4/22/23 10:48 AM, olcott wrote:
>>>>> On 4/22/2023 9:38 AM, Richard Damon wrote:
>>>>>> On 4/22/23 10:28 AM, olcott wrote:
>>>>>>> On 4/22/2023 6:17 AM, Richard Damon wrote:
>>>>>>>> On 4/21/23 11:40 PM, olcott wrote:
>>>>>>>>> On 4/21/2023 9:45 PM, Richard Damon wrote:
>>>>>>>>>> On 4/21/23 9:41 PM, olcott wrote:
>>>>>>>>>>> On 4/21/2023 7:49 PM, Richard Damon wrote:
>>>>>>>>>>>> On 4/21/23 8:33 PM, olcott wrote:
>>>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>>>>>>
>>>>>>>>>>>>> There exists a G such that G is logically equivalent to its
>>>>>>>>>>>>> own unprovability in F
>>>>>>>>>>>>>
>>>>>>>>>>>>> *If we assume that there is such a G in F that means that*
>>>>>>>>>>>>> G is true means there is no sequence of inference steps
>>>>>>>>>>>>> that satisfies G in F.
>>>>>>>>>>>>> G is false means there is a sequence of inference steps
>>>>>>>>>>>>> that satisfies G in F.
>>>>>>>>>>>>>
>>>>>>>>>>>>> *Thus the above G simply does not exist in F*
>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> So?
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> I finally learned enough model theory to correctly link
>>>>>>>>>>> provability to
>>>>>>>>>>> truth in the conventional model theory way.
>>>>>>>>>>
>>>>>>>>>> Doesn't seem so, you don't seem to understand the difference.
>>>>>>>>>> You seem to confuse Truth with Knowledge.
>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> I finally approximated {G asserts its own unprovability in F}
>>>>>>>>>>> using conventional math symbols in their conventional way.
>>>>>>>>>>
>>>>>>>>>> Except that isn't what G is, you only think that because you
>>>>>>>>>> can't actually understand even the outline of Godel's proof,
>>>>>>>>>> so you take pieces out of context.
>>>>>>>>>>
>>>>>>>>>> G never asserts its own unprovability.
>>>>>>>>>>
>>>>>>>>>> The statement that we now have a statement that asserts its
>>>>>>>>>> own unprovablity, as a simplification describing a statment
>>>>>>>>>> DERIVED from G, and that derivation happens in Meta-F, and is
>>>>>>>>>> about what can be proven in F.
>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>>> Since Godel's G isn't of that form, but only can be used to
>>>>>>>>>>>> derive a statment IN META-F that says that G is not provable
>>>>>>>>>>>> in F, your argument says nothing about Godel's G.
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> F ⊢ GF ↔ ¬ProvF (┌GF┐).
>>>>>>>>>>> https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
>>>>>>>>>>> I have finally created a G that is equivalent to
>>>>>>>>>>> Panu Raatikainen's SEP article.
>>>>>>>>>>
>>>>>>>>>> So?
>>>>>>>>>>
>>>>>>>>>> Did you read that article?
>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> Also, you don't understand what those terms mean, because G
>>>>>>>>>>>> being true doesn't mean there is no sequence of inference
>>>>>>>>>>>> steps that satisfies G in F, but there is no FINITE sequence
>>>>>>>>>>>> of inference steps that satisfies G in F.
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>>>>
>>>>>>>>>>> Because we can see that every finite or infinite sequence in
>>>>>>>>>>> F that
>>>>>>>>>>> satisfies the RHS of ↔ contradicts the LHS a powerful F can
>>>>>>>>>>> infer that G
>>>>>>>>>>> is utterly unsatisfiable even for infinite sequences in this
>>>>>>>>>>> more
>>>>>>>>>>> powerful F.
>>>>>>>>>>
>>>>>>>>>> Nope. Show the PROOF.
>>>>>>>>>>
>>>>>>>>>> You don't know HOW to do a proof, you can only do arguement.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>> There exists a G in F such that G is logically equivalent to
>>>>>>>>> its own unprovability in F
>>>>>>>>>
>>>>>>>>> A proof is any sequence of steps that shows that its conclusion
>>>>>>>>> is a
>>>>>>>>> necessary consequence of its premises.\
>>>>>>>>
>>>>>>>> Boy are you wrong.
>>>>>>>>
>>>>>>>> A proof is a FINITE sequence of steps that shows that a given
>>>>>>>> statement is a necessary consequence of the defined system.
>>>>>>>>
>>>>>>>> "Proof" doesn't have a  "Premise", it has a system.
>>>>>>>>
>>>>>>>> The statement may have conditions in it restricting when
>>>>>>>>
>>>>>>>>>
>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>> There exists a G in F such that G is logically equivalent to
>>>>>>>>> its own unprovability in F
>>>>>>>>>
>>>>>>>>> If G is true then there is no sequence of inference steps that
>>>>>>>>> satisfies G in F making G untrue.
>>>>>>>>
>>>>>>>> no FINITE sequence, making G UNPROVABLE, and there IS an
>>>>>>>> INFINITE sequence making it TRUE.
>>>>>>>>
>>>>>>>> This is possible.
>>>>>>>>
>>>>>>>>> If G is false then there is a sequence of inference steps that
>>>>>>>>> satisfies G in F making G true.
>>>>>>>>
>>>>>>>> If G is false, then there is a finite sequence proving G, which
>>>>>>>> forces G to be true, thus this is a contradiction.
>>>>>>>>
>>>>>>>>> Because the RHS of ↔ contradicts the LHS there is no such G in F.
>>>>>>>>> Thus the above G simply does not exist in F.
>>>>>>>>>
>>>>>>>>
>>>>>>>> Nope, because we can have an infinite sequence that isn't
>>>>>>>> finite, G can be True but not Provable.
>>>>>>>>
>>>>>>>
>>>>>>> If G is false and ↔ is true this makes the RHS false which
>>>>>>> negates the RHS making it say (G ⊢ F) which makes G true in F.
>>>>>>>
>>>>>>>
>>>>>>
>>>>>>
>>>>>> Right, G can't be false, but it can be True.
>>>>>>
>>>>>
>>>>> Thus ↔ cannot be satisfied thus no such G exists in F.
>>>>>
>>>>
>>>> Why do you say that?
>>>>
>>>> I don't think you know what you terms mean.
>>>>
>>>> There exists a G in F such that G is true if and only if G is
>>>> Unprovable.
>>>>
>>>
>>> Logical equality
>>> p q p ↔ q
>>> T T   T // G is true if and only if G is Unprovable.
>>> T F   F //
>>> F T   F //
>>> F F   T // G is false if and only if G is Provable.
>>> https://en.wikipedia.org/wiki/Truth_table#Logical_equality
>>>
>>> Row(1) There exists a G in F such that G is true if and only if G is
>>> unprovable in F making G unsatisfied thus untrue in F.
>>>
>>> Row(4) There exists a G in F such that G is false if and only if G is
>>> provable in F making G satisfied thus true in F.
>>>
>>> If either Row(1) or Row(4) are unsatisfied then ↔ is false.
>>
>> But if neither row values can ACTUALLY EXIST, then the equality is true.
>>
> If either Row(1) or Row(4) cannot have the same value for p and q
> (for whatever reason) then ↔ is unsatisfied and no such G exists in F.
>
So, you don't understand how truth tables work.

On 4/22/2023 11:36 AM, Richard Damon wrote:
> On 4/22/23 12:27 PM, olcott wrote:
>> On 4/22/2023 11:12 AM, Richard Damon wrote:
>>> On 4/22/23 11:39 AM, olcott wrote:
>>>> On 4/22/2023 9:57 AM, Richard Damon wrote:
>>>>> On 4/22/23 10:48 AM, olcott wrote:
>>>>>> On 4/22/2023 9:38 AM, Richard Damon wrote:
>>>>>>> On 4/22/23 10:28 AM, olcott wrote:
>>>>>>>> On 4/22/2023 6:17 AM, Richard Damon wrote:
>>>>>>>>> On 4/21/23 11:40 PM, olcott wrote:
>>>>>>>>>> On 4/21/2023 9:45 PM, Richard Damon wrote:
>>>>>>>>>>> On 4/21/23 9:41 PM, olcott wrote:
>>>>>>>>>>>> On 4/21/2023 7:49 PM, Richard Damon wrote:
>>>>>>>>>>>>> On 4/21/23 8:33 PM, olcott wrote:
>>>>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> There exists a G such that G is logically equivalent to
>>>>>>>>>>>>>> its own unprovability in F
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> *If we assume that there is such a G in F that means that*
>>>>>>>>>>>>>> G is true means there is no sequence of inference steps
>>>>>>>>>>>>>> that satisfies G in F.
>>>>>>>>>>>>>> G is false means there is a sequence of inference steps
>>>>>>>>>>>>>> that satisfies G in F.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> *Thus the above G simply does not exist in F*
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>> So?
>>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> I finally learned enough model theory to correctly link
>>>>>>>>>>>> provability to
>>>>>>>>>>>> truth in the conventional model theory way.
>>>>>>>>>>>
>>>>>>>>>>> Doesn't seem so, you don't seem to understand the difference.
>>>>>>>>>>> You seem to confuse Truth with Knowledge.
>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> I finally approximated {G asserts its own unprovability in F}
>>>>>>>>>>>> using conventional math symbols in their conventional way.
>>>>>>>>>>>
>>>>>>>>>>> Except that isn't what G is, you only think that because you
>>>>>>>>>>> can't actually understand even the outline of Godel's proof,
>>>>>>>>>>> so you take pieces out of context.
>>>>>>>>>>>
>>>>>>>>>>> G never asserts its own unprovability.
>>>>>>>>>>>
>>>>>>>>>>> The statement that we now have a statement that asserts its
>>>>>>>>>>> own unprovablity, as a simplification describing a statment
>>>>>>>>>>> DERIVED from G, and that derivation happens in Meta-F, and is
>>>>>>>>>>> about what can be proven in F.
>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>>> Since Godel's G isn't of that form, but only can be used to
>>>>>>>>>>>>> derive a statment IN META-F that says that G is not
>>>>>>>>>>>>> provable in F, your argument says nothing about Godel's G.
>>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> F ⊢ GF ↔ ¬ProvF (┌GF┐).
>>>>>>>>>>>> https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
>>>>>>>>>>>> I have finally created a G that is equivalent to
>>>>>>>>>>>> Panu Raatikainen's SEP article.
>>>>>>>>>>>
>>>>>>>>>>> So?
>>>>>>>>>>>
>>>>>>>>>>> Did you read that article?
>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>> Also, you don't understand what those terms mean, because G
>>>>>>>>>>>>> being true doesn't mean there is no sequence of inference
>>>>>>>>>>>>> steps that satisfies G in F, but there is no FINITE
>>>>>>>>>>>>> sequence of inference steps that satisfies G in F.
>>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>>>>>
>>>>>>>>>>>> Because we can see that every finite or infinite sequence in
>>>>>>>>>>>> F that
>>>>>>>>>>>> satisfies the RHS of ↔ contradicts the LHS a powerful F can
>>>>>>>>>>>> infer that G
>>>>>>>>>>>> is utterly unsatisfiable even for infinite sequences in this
>>>>>>>>>>>> more
>>>>>>>>>>>> powerful F.
>>>>>>>>>>>
>>>>>>>>>>> Nope. Show the PROOF.
>>>>>>>>>>>
>>>>>>>>>>> You don't know HOW to do a proof, you can only do arguement.
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>>> There exists a G in F such that G is logically equivalent to
>>>>>>>>>> its own unprovability in F
>>>>>>>>>>
>>>>>>>>>> A proof is any sequence of steps that shows that its
>>>>>>>>>> conclusion is a
>>>>>>>>>> necessary consequence of its premises.\
>>>>>>>>>
>>>>>>>>> Boy are you wrong.
>>>>>>>>>
>>>>>>>>> A proof is a FINITE sequence of steps that shows that a given
>>>>>>>>> statement is a necessary consequence of the defined system.
>>>>>>>>>
>>>>>>>>> "Proof" doesn't have a  "Premise", it has a system.
>>>>>>>>>
>>>>>>>>> The statement may have conditions in it restricting when
>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>>> There exists a G in F such that G is logically equivalent to
>>>>>>>>>> its own unprovability in F
>>>>>>>>>>
>>>>>>>>>> If G is true then there is no sequence of inference steps that
>>>>>>>>>> satisfies G in F making G untrue.
>>>>>>>>>
>>>>>>>>> no FINITE sequence, making G UNPROVABLE, and there IS an
>>>>>>>>> INFINITE sequence making it TRUE.
>>>>>>>>>
>>>>>>>>> This is possible.
>>>>>>>>>
>>>>>>>>>> If G is false then there is a sequence of inference steps that
>>>>>>>>>> satisfies G in F making G true.
>>>>>>>>>
>>>>>>>>> If G is false, then there is a finite sequence proving G, which
>>>>>>>>> forces G to be true, thus this is a contradiction.
>>>>>>>>>
>>>>>>>>>> Because the RHS of ↔ contradicts the LHS there is no such G in F.
>>>>>>>>>> Thus the above G simply does not exist in F.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> Nope, because we can have an infinite sequence that isn't
>>>>>>>>> finite, G can be True but not Provable.
>>>>>>>>>
>>>>>>>>
>>>>>>>> If G is false and ↔ is true this makes the RHS false which
>>>>>>>> negates the RHS making it say (G ⊢ F) which makes G true in F.
>>>>>>>>
>>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>> Right, G can't be false, but it can be True.
>>>>>>>
>>>>>>
>>>>>> Thus ↔ cannot be satisfied thus no such G exists in F.
>>>>>>
>>>>>
>>>>> Why do you say that?
>>>>>
>>>>> I don't think you know what you terms mean.
>>>>>
>>>>> There exists a G in F such that G is true if and only if G is
>>>>> Unprovable.
>>>>>
>>>>
>>>> Logical equality
>>>> p q p ↔ q
>>>> T T   T // G is true if and only if G is Unprovable.
>>>> T F   F //
>>>> F T   F //
>>>> F F   T // G is false if and only if G is Provable.
>>>> https://en.wikipedia.org/wiki/Truth_table#Logical_equality
>>>>
>>>> Row(1) There exists a G in F such that G is true if and only if G is
>>>> unprovable in F making G unsatisfied thus untrue in F.
>>>>
>>>> Row(4) There exists a G in F such that G is false if and only if G is
>>>> provable in F making G satisfied thus true in F.
>>>>
>>>> If either Row(1) or Row(4) are unsatisfied then ↔ is false.
>>>
>>> But if neither row values can ACTUALLY EXIST, then the equality is true.
>>>
>> If either Row(1) or Row(4) cannot have the same value for p and q
>> (for whatever reason) then ↔ is unsatisfied and no such G exists in F.
>>
> So, you don't understand how truth tables work.
>
> You don't need to have all the rows with true being possible, you need
> all the rows that are possible to be True.
>

On 4/22/23 12:45 PM, olcott wrote:
> On 4/22/2023 11:36 AM, Richard Damon wrote:
>> On 4/22/23 12:27 PM, olcott wrote:
>>> On 4/22/2023 11:12 AM, Richard Damon wrote:
>>>> On 4/22/23 11:39 AM, olcott wrote:
>>>>> On 4/22/2023 9:57 AM, Richard Damon wrote:
>>>>>> On 4/22/23 10:48 AM, olcott wrote:
>>>>>>> On 4/22/2023 9:38 AM, Richard Damon wrote:
>>>>>>>> On 4/22/23 10:28 AM, olcott wrote:
>>>>>>>>> On 4/22/2023 6:17 AM, Richard Damon wrote:
>>>>>>>>>> On 4/21/23 11:40 PM, olcott wrote:
>>>>>>>>>>> On 4/21/2023 9:45 PM, Richard Damon wrote:
>>>>>>>>>>>> On 4/21/23 9:41 PM, olcott wrote:
>>>>>>>>>>>>> On 4/21/2023 7:49 PM, Richard Damon wrote:
>>>>>>>>>>>>>> On 4/21/23 8:33 PM, olcott wrote:
>>>>>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> There exists a G such that G is logically equivalent to
>>>>>>>>>>>>>>> its own unprovability in F
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> *If we assume that there is such a G in F that means that*
>>>>>>>>>>>>>>> G is true means there is no sequence of inference steps
>>>>>>>>>>>>>>> that satisfies G in F.
>>>>>>>>>>>>>>> G is false means there is a sequence of inference steps
>>>>>>>>>>>>>>> that satisfies G in F.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> *Thus the above G simply does not exist in F*
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> So?
>>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>> I finally learned enough model theory to correctly link
>>>>>>>>>>>>> provability to
>>>>>>>>>>>>> truth in the conventional model theory way.
>>>>>>>>>>>>
>>>>>>>>>>>> Doesn't seem so, you don't seem to understand the
>>>>>>>>>>>> difference. You seem to confuse Truth with Knowledge.
>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>> I finally approximated {G asserts its own unprovability in F}
>>>>>>>>>>>>> using conventional math symbols in their conventional way.
>>>>>>>>>>>>
>>>>>>>>>>>> Except that isn't what G is, you only think that because you
>>>>>>>>>>>> can't actually understand even the outline of Godel's proof,
>>>>>>>>>>>> so you take pieces out of context.
>>>>>>>>>>>>
>>>>>>>>>>>> G never asserts its own unprovability.
>>>>>>>>>>>>
>>>>>>>>>>>> The statement that we now have a statement that asserts its
>>>>>>>>>>>> own unprovablity, as a simplification describing a statment
>>>>>>>>>>>> DERIVED from G, and that derivation happens in Meta-F, and
>>>>>>>>>>>> is about what can be proven in F.
>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>>> Since Godel's G isn't of that form, but only can be used
>>>>>>>>>>>>>> to derive a statment IN META-F that says that G is not
>>>>>>>>>>>>>> provable in F, your argument says nothing about Godel's G.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>> F ⊢ GF ↔ ¬ProvF (┌GF┐).
>>>>>>>>>>>>> https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
>>>>>>>>>>>>> I have finally created a G that is equivalent to
>>>>>>>>>>>>> Panu Raatikainen's SEP article.
>>>>>>>>>>>>
>>>>>>>>>>>> So?
>>>>>>>>>>>>
>>>>>>>>>>>> Did you read that article?
>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Also, you don't understand what those terms mean, because
>>>>>>>>>>>>>> G being true doesn't mean there is no sequence of
>>>>>>>>>>>>>> inference steps that satisfies G in F, but there is no
>>>>>>>>>>>>>> FINITE sequence of inference steps that satisfies G in F.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>>>>>>
>>>>>>>>>>>>> Because we can see that every finite or infinite sequence
>>>>>>>>>>>>> in F that
>>>>>>>>>>>>> satisfies the RHS of ↔ contradicts the LHS a powerful F can
>>>>>>>>>>>>> infer that G
>>>>>>>>>>>>> is utterly unsatisfiable even for infinite sequences in
>>>>>>>>>>>>> this more
>>>>>>>>>>>>> powerful F.
>>>>>>>>>>>>
>>>>>>>>>>>> Nope. Show the PROOF.
>>>>>>>>>>>>
>>>>>>>>>>>> You don't know HOW to do a proof, you can only do arguement.
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>>>> There exists a G in F such that G is logically equivalent to
>>>>>>>>>>> its own unprovability in F
>>>>>>>>>>>
>>>>>>>>>>> A proof is any sequence of steps that shows that its
>>>>>>>>>>> conclusion is a
>>>>>>>>>>> necessary consequence of its premises.\
>>>>>>>>>>
>>>>>>>>>> Boy are you wrong.
>>>>>>>>>>
>>>>>>>>>> A proof is a FINITE sequence of steps that shows that a given
>>>>>>>>>> statement is a necessary consequence of the defined system.
>>>>>>>>>>
>>>>>>>>>> "Proof" doesn't have a  "Premise", it has a system.
>>>>>>>>>>
>>>>>>>>>> The statement may have conditions in it restricting when
>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>>>> There exists a G in F such that G is logically equivalent to
>>>>>>>>>>> its own unprovability in F
>>>>>>>>>>>
>>>>>>>>>>> If G is true then there is no sequence of inference steps
>>>>>>>>>>> that satisfies G in F making G untrue.
>>>>>>>>>>
>>>>>>>>>> no FINITE sequence, making G UNPROVABLE, and there IS an
>>>>>>>>>> INFINITE sequence making it TRUE.
>>>>>>>>>>
>>>>>>>>>> This is possible.
>>>>>>>>>>
>>>>>>>>>>> If G is false then there is a sequence of inference steps
>>>>>>>>>>> that satisfies G in F making G true.
>>>>>>>>>>
>>>>>>>>>> If G is false, then there is a finite sequence proving G,
>>>>>>>>>> which forces G to be true, thus this is a contradiction.
>>>>>>>>>>
>>>>>>>>>>> Because the RHS of ↔ contradicts the LHS there is no such G
>>>>>>>>>>> in F.
>>>>>>>>>>> Thus the above G simply does not exist in F.
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> Nope, because we can have an infinite sequence that isn't
>>>>>>>>>> finite, G can be True but not Provable.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> If G is false and ↔ is true this makes the RHS false which
>>>>>>>>> negates the RHS making it say (G ⊢ F) which makes G true in F.
>>>>>>>>>
>>>>>>>>>
>>>>>>>>
>>>>>>>>
>>>>>>>> Right, G can't be false, but it can be True.
>>>>>>>>
>>>>>>>
>>>>>>> Thus ↔ cannot be satisfied thus no such G exists in F.
>>>>>>>
>>>>>>
>>>>>> Why do you say that?
>>>>>>
>>>>>> I don't think you know what you terms mean.
>>>>>>
>>>>>> There exists a G in F such that G is true if and only if G is
>>>>>> Unprovable.
>>>>>>
>>>>>
>>>>> Logical equality
>>>>> p q p ↔ q
>>>>> T T   T // G is true if and only if G is Unprovable.
>>>>> T F   F //
>>>>> F T   F //
>>>>> F F   T // G is false if and only if G is Provable.
>>>>> https://en.wikipedia.org/wiki/Truth_table#Logical_equality
>>>>>
>>>>> Row(1) There exists a G in F such that G is true if and only if G is
>>>>> unprovable in F making G unsatisfied thus untrue in F.
>>>>>
>>>>> Row(4) There exists a G in F such that G is false if and only if G is
>>>>> provable in F making G satisfied thus true in F.
>>>>>
>>>>> If either Row(1) or Row(4) are unsatisfied then ↔ is false.
>>>>
>>>> But if neither row values can ACTUALLY EXIST, then the equality is
>>>> true.
>>>>
>>> If either Row(1) or Row(4) cannot have the same value for p and q
>>> (for whatever reason) then ↔ is unsatisfied and no such G exists in F.
>>>
>> So, you don't understand how truth tables work.
>>
>> You don't need to have all the rows with true being possible, you need
>> all the rows that are possible to be True.
>>
>
> To the best of my knowledge
> ↔ is also known as logical equivalence meaning that the LHS and the RHS
> must always have the same truth value or ↔ is not true.
>

On 4/22/2023 11:56 AM, Richard Damon wrote:
> On 4/22/23 12:45 PM, olcott wrote:
>> On 4/22/2023 11:36 AM, Richard Damon wrote:
>>> On 4/22/23 12:27 PM, olcott wrote:
>>>> On 4/22/2023 11:12 AM, Richard Damon wrote:
>>>>> On 4/22/23 11:39 AM, olcott wrote:
>>>>>> On 4/22/2023 9:57 AM, Richard Damon wrote:
>>>>>>> On 4/22/23 10:48 AM, olcott wrote:
>>>>>>>> On 4/22/2023 9:38 AM, Richard Damon wrote:
>>>>>>>>> On 4/22/23 10:28 AM, olcott wrote:
>>>>>>>>>> On 4/22/2023 6:17 AM, Richard Damon wrote:
>>>>>>>>>>> On 4/21/23 11:40 PM, olcott wrote:
>>>>>>>>>>>> On 4/21/2023 9:45 PM, Richard Damon wrote:
>>>>>>>>>>>>> On 4/21/23 9:41 PM, olcott wrote:
>>>>>>>>>>>>>> On 4/21/2023 7:49 PM, Richard Damon wrote:
>>>>>>>>>>>>>>> On 4/21/23 8:33 PM, olcott wrote:
>>>>>>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> There exists a G such that G is logically equivalent to
>>>>>>>>>>>>>>>> its own unprovability in F
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> *If we assume that there is such a G in F that means that*
>>>>>>>>>>>>>>>> G is true means there is no sequence of inference steps
>>>>>>>>>>>>>>>> that satisfies G in F.
>>>>>>>>>>>>>>>> G is false means there is a sequence of inference steps
>>>>>>>>>>>>>>>> that satisfies G in F.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> *Thus the above G simply does not exist in F*
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> So?
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> I finally learned enough model theory to correctly link
>>>>>>>>>>>>>> provability to
>>>>>>>>>>>>>> truth in the conventional model theory way.
>>>>>>>>>>>>>
>>>>>>>>>>>>> Doesn't seem so, you don't seem to understand the
>>>>>>>>>>>>> difference. You seem to confuse Truth with Knowledge.
>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> I finally approximated {G asserts its own unprovability in F}
>>>>>>>>>>>>>> using conventional math symbols in their conventional way.
>>>>>>>>>>>>>
>>>>>>>>>>>>> Except that isn't what G is, you only think that because
>>>>>>>>>>>>> you can't actually understand even the outline of Godel's
>>>>>>>>>>>>> proof, so you take pieces out of context.
>>>>>>>>>>>>>
>>>>>>>>>>>>> G never asserts its own unprovability.
>>>>>>>>>>>>>
>>>>>>>>>>>>> The statement that we now have a statement that asserts its
>>>>>>>>>>>>> own unprovablity, as a simplification describing a statment
>>>>>>>>>>>>> DERIVED from G, and that derivation happens in Meta-F, and
>>>>>>>>>>>>> is about what can be proven in F.
>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Since Godel's G isn't of that form, but only can be used
>>>>>>>>>>>>>>> to derive a statment IN META-F that says that G is not
>>>>>>>>>>>>>>> provable in F, your argument says nothing about Godel's G.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> F ⊢ GF ↔ ¬ProvF (┌GF┐).
>>>>>>>>>>>>>> https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
>>>>>>>>>>>>>> I have finally created a G that is equivalent to
>>>>>>>>>>>>>> Panu Raatikainen's SEP article.
>>>>>>>>>>>>>
>>>>>>>>>>>>> So?
>>>>>>>>>>>>>
>>>>>>>>>>>>> Did you read that article?
>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Also, you don't understand what those terms mean, because
>>>>>>>>>>>>>>> G being true doesn't mean there is no sequence of
>>>>>>>>>>>>>>> inference steps that satisfies G in F, but there is no
>>>>>>>>>>>>>>> FINITE sequence of inference steps that satisfies G in F.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Because we can see that every finite or infinite sequence
>>>>>>>>>>>>>> in F that
>>>>>>>>>>>>>> satisfies the RHS of ↔ contradicts the LHS a powerful F
>>>>>>>>>>>>>> can infer that G
>>>>>>>>>>>>>> is utterly unsatisfiable even for infinite sequences in
>>>>>>>>>>>>>> this more
>>>>>>>>>>>>>> powerful F.
>>>>>>>>>>>>>
>>>>>>>>>>>>> Nope. Show the PROOF.
>>>>>>>>>>>>>
>>>>>>>>>>>>> You don't know HOW to do a proof, you can only do arguement.
>>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>>>>> There exists a G in F such that G is logically equivalent to
>>>>>>>>>>>> its own unprovability in F
>>>>>>>>>>>>
>>>>>>>>>>>> A proof is any sequence of steps that shows that its
>>>>>>>>>>>> conclusion is a
>>>>>>>>>>>> necessary consequence of its premises.\
>>>>>>>>>>>
>>>>>>>>>>> Boy are you wrong.
>>>>>>>>>>>
>>>>>>>>>>> A proof is a FINITE sequence of steps that shows that a given
>>>>>>>>>>> statement is a necessary consequence of the defined system.
>>>>>>>>>>>
>>>>>>>>>>> "Proof" doesn't have a  "Premise", it has a system.
>>>>>>>>>>>
>>>>>>>>>>> The statement may have conditions in it restricting when
>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>>>>> There exists a G in F such that G is logically equivalent to
>>>>>>>>>>>> its own unprovability in F
>>>>>>>>>>>>
>>>>>>>>>>>> If G is true then there is no sequence of inference steps
>>>>>>>>>>>> that satisfies G in F making G untrue.
>>>>>>>>>>>
>>>>>>>>>>> no FINITE sequence, making G UNPROVABLE, and there IS an
>>>>>>>>>>> INFINITE sequence making it TRUE.
>>>>>>>>>>>
>>>>>>>>>>> This is possible.
>>>>>>>>>>>
>>>>>>>>>>>> If G is false then there is a sequence of inference steps
>>>>>>>>>>>> that satisfies G in F making G true.
>>>>>>>>>>>
>>>>>>>>>>> If G is false, then there is a finite sequence proving G,
>>>>>>>>>>> which forces G to be true, thus this is a contradiction.
>>>>>>>>>>>
>>>>>>>>>>>> Because the RHS of ↔ contradicts the LHS there is no such G
>>>>>>>>>>>> in F.
>>>>>>>>>>>> Thus the above G simply does not exist in F.
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> Nope, because we can have an infinite sequence that isn't
>>>>>>>>>>> finite, G can be True but not Provable.
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> If G is false and ↔ is true this makes the RHS false which
>>>>>>>>>> negates the RHS making it say (G ⊢ F) which makes G true in F.
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>>
>>>>>>>>> Right, G can't be false, but it can be True.
>>>>>>>>>
>>>>>>>>
>>>>>>>> Thus ↔ cannot be satisfied thus no such G exists in F.
>>>>>>>>
>>>>>>>
>>>>>>> Why do you say that?
>>>>>>>
>>>>>>> I don't think you know what you terms mean.
>>>>>>>
>>>>>>> There exists a G in F such that G is true if and only if G is
>>>>>>> Unprovable.
>>>>>>>
>>>>>>
>>>>>> Logical equality
>>>>>> p q p ↔ q
>>>>>> T T   T // G is true if and only if G is Unprovable.
>>>>>> T F   F //
>>>>>> F T   F //
>>>>>> F F   T // G is false if and only if G is Provable.
>>>>>> https://en.wikipedia.org/wiki/Truth_table#Logical_equality
>>>>>>
>>>>>> Row(1) There exists a G in F such that G is true if and only if G is
>>>>>> unprovable in F making G unsatisfied thus untrue in F.
>>>>>>
>>>>>> Row(4) There exists a G in F such that G is false if and only if G is
>>>>>> provable in F making G satisfied thus true in F.
>>>>>>
>>>>>> If either Row(1) or Row(4) are unsatisfied then ↔ is false.
>>>>>
>>>>> But if neither row values can ACTUALLY EXIST, then the equality is
>>>>> true.
>>>>>
>>>> If either Row(1) or Row(4) cannot have the same value for p and q
>>>> (for whatever reason) then ↔ is unsatisfied and no such G exists in F.
>>>>
>>> So, you don't understand how truth tables work.
>>>
>>> You don't need to have all the rows with true being possible, you
>>> need all the rows that are possible to be True.
>>>
>>
>> To the best of my knowledge
>> ↔ is also known as logical equivalence meaning that the LHS and the RHS
>> must always have the same truth value or ↔ is not true.
>>
>
> Right, and for that statement, the actual G found in F, the ONLY values
> that happen is G is ALWAYS true, an Unprovable is always true.
>
> Thus the equivalence is always true.
I don't think that is the way that it works.
We must assume that the RHS is true and see how that effects the LHS
We must assume that the RHS is false and see how that effects the LHS
((True(RHS) → True(LHS)) ∧ (False(RHS) → False(LHS))) ≡ (RHS ↔ LHS)
False(RHS) → True(LHS) refutes (RHS ↔ LHS)

On 4/22/23 1:13 PM, olcott wrote:
> On 4/22/2023 11:56 AM, Richard Damon wrote:
>> On 4/22/23 12:45 PM, olcott wrote:
>>> On 4/22/2023 11:36 AM, Richard Damon wrote:
>>>> On 4/22/23 12:27 PM, olcott wrote:
>>>>> On 4/22/2023 11:12 AM, Richard Damon wrote:
>>>>>> On 4/22/23 11:39 AM, olcott wrote:
>>>>>>> On 4/22/2023 9:57 AM, Richard Damon wrote:
>>>>>>>> On 4/22/23 10:48 AM, olcott wrote:
>>>>>>>>> On 4/22/2023 9:38 AM, Richard Damon wrote:
>>>>>>>>>> On 4/22/23 10:28 AM, olcott wrote:
>>>>>>>>>>> On 4/22/2023 6:17 AM, Richard Damon wrote:
>>>>>>>>>>>> On 4/21/23 11:40 PM, olcott wrote:
>>>>>>>>>>>>> On 4/21/2023 9:45 PM, Richard Damon wrote:
>>>>>>>>>>>>>> On 4/21/23 9:41 PM, olcott wrote:
>>>>>>>>>>>>>>> On 4/21/2023 7:49 PM, Richard Damon wrote:
>>>>>>>>>>>>>>>> On 4/21/23 8:33 PM, olcott wrote:
>>>>>>>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> There exists a G such that G is logically equivalent to
>>>>>>>>>>>>>>>>> its own unprovability in F
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> *If we assume that there is such a G in F that means that*
>>>>>>>>>>>>>>>>> G is true means there is no sequence of inference steps
>>>>>>>>>>>>>>>>> that satisfies G in F.
>>>>>>>>>>>>>>>>> G is false means there is a sequence of inference steps
>>>>>>>>>>>>>>>>> that satisfies G in F.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> *Thus the above G simply does not exist in F*
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> So?
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> I finally learned enough model theory to correctly link
>>>>>>>>>>>>>>> provability to
>>>>>>>>>>>>>>> truth in the conventional model theory way.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Doesn't seem so, you don't seem to understand the
>>>>>>>>>>>>>> difference. You seem to confuse Truth with Knowledge.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> I finally approximated {G asserts its own unprovability
>>>>>>>>>>>>>>> in F}
>>>>>>>>>>>>>>> using conventional math symbols in their conventional way.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Except that isn't what G is, you only think that because
>>>>>>>>>>>>>> you can't actually understand even the outline of Godel's
>>>>>>>>>>>>>> proof, so you take pieces out of context.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> G never asserts its own unprovability.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> The statement that we now have a statement that asserts
>>>>>>>>>>>>>> its own unprovablity, as a simplification describing a
>>>>>>>>>>>>>> statment DERIVED from G, and that derivation happens in
>>>>>>>>>>>>>> Meta-F, and is about what can be proven in F.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> Since Godel's G isn't of that form, but only can be used
>>>>>>>>>>>>>>>> to derive a statment IN META-F that says that G is not
>>>>>>>>>>>>>>>> provable in F, your argument says nothing about Godel's G.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> F ⊢ GF ↔ ¬ProvF (┌GF┐).
>>>>>>>>>>>>>>> https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
>>>>>>>>>>>>>>> I have finally created a G that is equivalent to
>>>>>>>>>>>>>>> Panu Raatikainen's SEP article.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> So?
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Did you read that article?
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> Also, you don't understand what those terms mean,
>>>>>>>>>>>>>>>> because G being true doesn't mean there is no sequence
>>>>>>>>>>>>>>>> of inference steps that satisfies G in F, but there is
>>>>>>>>>>>>>>>> no FINITE sequence of inference steps that satisfies G
>>>>>>>>>>>>>>>> in F.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Because we can see that every finite or infinite sequence
>>>>>>>>>>>>>>> in F that
>>>>>>>>>>>>>>> satisfies the RHS of ↔ contradicts the LHS a powerful F
>>>>>>>>>>>>>>> can infer that G
>>>>>>>>>>>>>>> is utterly unsatisfiable even for infinite sequences in
>>>>>>>>>>>>>>> this more
>>>>>>>>>>>>>>> powerful F.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Nope. Show the PROOF.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> You don't know HOW to do a proof, you can only do arguement.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>>>>>> There exists a G in F such that G is logically equivalent
>>>>>>>>>>>>> to its own unprovability in F
>>>>>>>>>>>>>
>>>>>>>>>>>>> A proof is any sequence of steps that shows that its
>>>>>>>>>>>>> conclusion is a
>>>>>>>>>>>>> necessary consequence of its premises.\
>>>>>>>>>>>>
>>>>>>>>>>>> Boy are you wrong.
>>>>>>>>>>>>
>>>>>>>>>>>> A proof is a FINITE sequence of steps that shows that a
>>>>>>>>>>>> given statement is a necessary consequence of the defined
>>>>>>>>>>>> system.
>>>>>>>>>>>>
>>>>>>>>>>>> "Proof" doesn't have a  "Premise", it has a system.
>>>>>>>>>>>>
>>>>>>>>>>>> The statement may have conditions in it restricting when
>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>>>>>> There exists a G in F such that G is logically equivalent
>>>>>>>>>>>>> to its own unprovability in F
>>>>>>>>>>>>>
>>>>>>>>>>>>> If G is true then there is no sequence of inference steps
>>>>>>>>>>>>> that satisfies G in F making G untrue.
>>>>>>>>>>>>
>>>>>>>>>>>> no FINITE sequence, making G UNPROVABLE, and there IS an
>>>>>>>>>>>> INFINITE sequence making it TRUE.
>>>>>>>>>>>>
>>>>>>>>>>>> This is possible.
>>>>>>>>>>>>
>>>>>>>>>>>>> If G is false then there is a sequence of inference steps
>>>>>>>>>>>>> that satisfies G in F making G true.
>>>>>>>>>>>>
>>>>>>>>>>>> If G is false, then there is a finite sequence proving G,
>>>>>>>>>>>> which forces G to be true, thus this is a contradiction.
>>>>>>>>>>>>
>>>>>>>>>>>>> Because the RHS of ↔ contradicts the LHS there is no such G
>>>>>>>>>>>>> in F.
>>>>>>>>>>>>> Thus the above G simply does not exist in F.
>>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> Nope, because we can have an infinite sequence that isn't
>>>>>>>>>>>> finite, G can be True but not Provable.
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> If G is false and ↔ is true this makes the RHS false which
>>>>>>>>>>> negates the RHS making it say (G ⊢ F) which makes G true in F.
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> Right, G can't be false, but it can be True.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> Thus ↔ cannot be satisfied thus no such G exists in F.
>>>>>>>>>
>>>>>>>>
>>>>>>>> Why do you say that?
>>>>>>>>
>>>>>>>> I don't think you know what you terms mean.
>>>>>>>>
>>>>>>>> There exists a G in F such that G is true if and only if G is
>>>>>>>> Unprovable.
>>>>>>>>
>>>>>>>
>>>>>>> Logical equality
>>>>>>> p q p ↔ q
>>>>>>> T T   T // G is true if and only if G is Unprovable.
>>>>>>> T F   F //
>>>>>>> F T   F //
>>>>>>> F F   T // G is false if and only if G is Provable.
>>>>>>> https://en.wikipedia.org/wiki/Truth_table#Logical_equality
>>>>>>>
>>>>>>> Row(1) There exists a G in F such that G is true if and only if G is
>>>>>>> unprovable in F making G unsatisfied thus untrue in F.
>>>>>>>
>>>>>>> Row(4) There exists a G in F such that G is false if and only if
>>>>>>> G is
>>>>>>> provable in F making G satisfied thus true in F.
>>>>>>>
>>>>>>> If either Row(1) or Row(4) are unsatisfied then ↔ is false.
>>>>>>
>>>>>> But if neither row values can ACTUALLY EXIST, then the equality is
>>>>>> true.
>>>>>>
>>>>> If either Row(1) or Row(4) cannot have the same value for p and q
>>>>> (for whatever reason) then ↔ is unsatisfied and no such G exists in F.
>>>>>
>>>> So, you don't understand how truth tables work.
>>>>
>>>> You don't need to have all the rows with true being possible, you
>>>> need all the rows that are possible to be True.
>>>>
>>>
>>> To the best of my knowledge
>>> ↔ is also known as logical equivalence meaning that the LHS and the RHS
>>> must always have the same truth value or ↔ is not true.
>>>
>>
>> Right, and for that statement, the actual G found in F, the ONLY
>> values that happen is G is ALWAYS true, an Unprovable is always true.
>>
>> Thus the equivalence is always true.
> I don't think that is the way that it works.
> We must assume that the RHS is true and see how that effects the LHS
> We must assume that the RHS is false and see how that effects the LHS
> ((True(RHS) → True(LHS)) ∧ (False(RHS) → False(LHS))) ≡ (RHS ↔ LHS)
> False(RHS) → True(LHS) refutes (RHS ↔ LHS)
>

On 4/22/2023 1:01 PM, Richard Damon wrote:
> On 4/22/23 1:13 PM, olcott wrote:
>> On 4/22/2023 11:56 AM, Richard Damon wrote:
>>> On 4/22/23 12:45 PM, olcott wrote:
>>>> On 4/22/2023 11:36 AM, Richard Damon wrote:
>>>>> On 4/22/23 12:27 PM, olcott wrote:
>>>>>> On 4/22/2023 11:12 AM, Richard Damon wrote:
>>>>>>> On 4/22/23 11:39 AM, olcott wrote:
>>>>>>>> On 4/22/2023 9:57 AM, Richard Damon wrote:
>>>>>>>>> On 4/22/23 10:48 AM, olcott wrote:
>>>>>>>>>> On 4/22/2023 9:38 AM, Richard Damon wrote:
>>>>>>>>>>> On 4/22/23 10:28 AM, olcott wrote:
>>>>>>>>>>>> On 4/22/2023 6:17 AM, Richard Damon wrote:
>>>>>>>>>>>>> On 4/21/23 11:40 PM, olcott wrote:
>>>>>>>>>>>>>> On 4/21/2023 9:45 PM, Richard Damon wrote:
>>>>>>>>>>>>>>> On 4/21/23 9:41 PM, olcott wrote:
>>>>>>>>>>>>>>>> On 4/21/2023 7:49 PM, Richard Damon wrote:
>>>>>>>>>>>>>>>>> On 4/21/23 8:33 PM, olcott wrote:
>>>>>>>>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> There exists a G such that G is logically equivalent
>>>>>>>>>>>>>>>>>> to its own unprovability in F
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> *If we assume that there is such a G in F that means
>>>>>>>>>>>>>>>>>> that*
>>>>>>>>>>>>>>>>>> G is true means there is no sequence of inference
>>>>>>>>>>>>>>>>>> steps that satisfies G in F.
>>>>>>>>>>>>>>>>>> G is false means there is a sequence of inference
>>>>>>>>>>>>>>>>>> steps that satisfies G in F.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> *Thus the above G simply does not exist in F*
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> So?
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> I finally learned enough model theory to correctly link
>>>>>>>>>>>>>>>> provability to
>>>>>>>>>>>>>>>> truth in the conventional model theory way.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Doesn't seem so, you don't seem to understand the
>>>>>>>>>>>>>>> difference. You seem to confuse Truth with Knowledge.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> I finally approximated {G asserts its own unprovability
>>>>>>>>>>>>>>>> in F}
>>>>>>>>>>>>>>>> using conventional math symbols in their conventional way.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Except that isn't what G is, you only think that because
>>>>>>>>>>>>>>> you can't actually understand even the outline of Godel's
>>>>>>>>>>>>>>> proof, so you take pieces out of context.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> G never asserts its own unprovability.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> The statement that we now have a statement that asserts
>>>>>>>>>>>>>>> its own unprovablity, as a simplification describing a
>>>>>>>>>>>>>>> statment DERIVED from G, and that derivation happens in
>>>>>>>>>>>>>>> Meta-F, and is about what can be proven in F.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> Since Godel's G isn't of that form, but only can be
>>>>>>>>>>>>>>>>> used to derive a statment IN META-F that says that G is
>>>>>>>>>>>>>>>>> not provable in F, your argument says nothing about
>>>>>>>>>>>>>>>>> Godel's G.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> F ⊢ GF ↔ ¬ProvF (┌GF┐).
>>>>>>>>>>>>>>>> https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
>>>>>>>>>>>>>>>> I have finally created a G that is equivalent to
>>>>>>>>>>>>>>>> Panu Raatikainen's SEP article.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> So?
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Did you read that article?
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> Also, you don't understand what those terms mean,
>>>>>>>>>>>>>>>>> because G being true doesn't mean there is no sequence
>>>>>>>>>>>>>>>>> of inference steps that satisfies G in F, but there is
>>>>>>>>>>>>>>>>> no FINITE sequence of inference steps that satisfies G
>>>>>>>>>>>>>>>>> in F.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> Because we can see that every finite or infinite
>>>>>>>>>>>>>>>> sequence in F that
>>>>>>>>>>>>>>>> satisfies the RHS of ↔ contradicts the LHS a powerful F
>>>>>>>>>>>>>>>> can infer that G
>>>>>>>>>>>>>>>> is utterly unsatisfiable even for infinite sequences in
>>>>>>>>>>>>>>>> this more
>>>>>>>>>>>>>>>> powerful F.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Nope. Show the PROOF.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> You don't know HOW to do a proof, you can only do arguement.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>>>>>>> There exists a G in F such that G is logically equivalent
>>>>>>>>>>>>>> to its own unprovability in F
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> A proof is any sequence of steps that shows that its
>>>>>>>>>>>>>> conclusion is a
>>>>>>>>>>>>>> necessary consequence of its premises.\
>>>>>>>>>>>>>
>>>>>>>>>>>>> Boy are you wrong.
>>>>>>>>>>>>>
>>>>>>>>>>>>> A proof is a FINITE sequence of steps that shows that a
>>>>>>>>>>>>> given statement is a necessary consequence of the defined
>>>>>>>>>>>>> system.
>>>>>>>>>>>>>
>>>>>>>>>>>>> "Proof" doesn't have a  "Premise", it has a system.
>>>>>>>>>>>>>
>>>>>>>>>>>>> The statement may have conditions in it restricting when
>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>>>>>>> There exists a G in F such that G is logically equivalent
>>>>>>>>>>>>>> to its own unprovability in F
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> If G is true then there is no sequence of inference steps
>>>>>>>>>>>>>> that satisfies G in F making G untrue.
>>>>>>>>>>>>>
>>>>>>>>>>>>> no FINITE sequence, making G UNPROVABLE, and there IS an
>>>>>>>>>>>>> INFINITE sequence making it TRUE.
>>>>>>>>>>>>>
>>>>>>>>>>>>> This is possible.
>>>>>>>>>>>>>
>>>>>>>>>>>>>> If G is false then there is a sequence of inference steps
>>>>>>>>>>>>>> that satisfies G in F making G true.
>>>>>>>>>>>>>
>>>>>>>>>>>>> If G is false, then there is a finite sequence proving G,
>>>>>>>>>>>>> which forces G to be true, thus this is a contradiction.
>>>>>>>>>>>>>
>>>>>>>>>>>>>> Because the RHS of ↔ contradicts the LHS there is no such
>>>>>>>>>>>>>> G in F.
>>>>>>>>>>>>>> Thus the above G simply does not exist in F.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>> Nope, because we can have an infinite sequence that isn't
>>>>>>>>>>>>> finite, G can be True but not Provable.
>>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> If G is false and ↔ is true this makes the RHS false which
>>>>>>>>>>>> negates the RHS making it say (G ⊢ F) which makes G true in F.
>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> Right, G can't be false, but it can be True.
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> Thus ↔ cannot be satisfied thus no such G exists in F.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> Why do you say that?
>>>>>>>>>
>>>>>>>>> I don't think you know what you terms mean.
>>>>>>>>>
>>>>>>>>> There exists a G in F such that G is true if and only if G is
>>>>>>>>> Unprovable.
>>>>>>>>>
>>>>>>>>
>>>>>>>> Logical equality
>>>>>>>> p q p ↔ q
>>>>>>>> T T   T // G is true if and only if G is Unprovable.
>>>>>>>> T F   F //
>>>>>>>> F T   F //
>>>>>>>> F F   T // G is false if and only if G is Provable.
>>>>>>>> https://en.wikipedia.org/wiki/Truth_table#Logical_equality
>>>>>>>>
>>>>>>>> Row(1) There exists a G in F such that G is true if and only if
>>>>>>>> G is
>>>>>>>> unprovable in F making G unsatisfied thus untrue in F.
>>>>>>>>
>>>>>>>> Row(4) There exists a G in F such that G is false if and only if
>>>>>>>> G is
>>>>>>>> provable in F making G satisfied thus true in F.
>>>>>>>>
>>>>>>>> If either Row(1) or Row(4) are unsatisfied then ↔ is false.
>>>>>>>
>>>>>>> But if neither row values can ACTUALLY EXIST, then the equality
>>>>>>> is true.
>>>>>>>
>>>>>> If either Row(1) or Row(4) cannot have the same value for p and q
>>>>>> (for whatever reason) then ↔ is unsatisfied and no such G exists
>>>>>> in F.
>>>>>>
>>>>> So, you don't understand how truth tables work.
>>>>>
>>>>> You don't need to have all the rows with true being possible, you
>>>>> need all the rows that are possible to be True.
>>>>>
>>>>
>>>> To the best of my knowledge
>>>> ↔ is also known as logical equivalence meaning that the LHS and the RHS
>>>> must always have the same truth value or ↔ is not true.
>>>>
>>>
>>> Right, and for that statement, the actual G found in F, the ONLY
>>> values that happen is G is ALWAYS true, an Unprovable is always true.
>>>
>>> Thus the equivalence is always true.
>> I don't think that is the way that it works.
>> We must assume that the RHS is true and see how that effects the LHS
>> We must assume that the RHS is false and see how that effects the LHS
>> ((True(RHS) → True(LHS)) ∧ (False(RHS) → False(LHS))) ≡ (RHS ↔ LHS)
>> False(RHS) → True(LHS) refutes (RHS ↔ LHS)
>>
>
> Nope, that isn't how it works.
>
> Can you show me something that says that is how it works?
>

On 4/22/23 1:13 PM, olcott wrote:
>> Right, and for that statement, the actual G found in F, the ONLY
>> values that happen is G is ALWAYS true, an Unprovable is always true.
>>
>> Thus the equivalence is always true.
> I don't think that is the way that it works.
> We must assume that the RHS is true and see how that effects the LHS
> We must assume that the RHS is false and see how that effects the LHS
> ((True(RHS) → True(LHS)) ∧ (False(RHS) → False(LHS))) ≡ (RHS ↔ LHS)
> False(RHS) → True(LHS) refutes (RHS ↔ LHS)

And the other thing you are forgetting is that you are looking at this
in a CONDITION.

Remember, you were asking:

∃G ∈ F (G ↔ (G ⊬ F))

So, G ↔ (G ⊬ F) doesn't need to be true for any arbitrary statement G,
but we are asking if there exist a SPECIFIC G, for which it holds. It is
stating that There does exact AT LEAST ONE G, that is in F, that the
relationship hold.

Godel's G, fits the bill, and since it doesn't depend on models within
F, but is a fundamental property of F itself, we have a case that the
statements "G" and "G ⊬ F" are simple binary values, not something that
varies over the model space of F, and both are True.

Yes, for a lot of other statements, you will find some that are not
provable, because they are false, and some that are true, but provable.
Either of those conditions show that THAT statement doesn't meet the
condition.

You will NOT find any statements that are False, but Provable, so line 4
of you table will NEVER actually be used, due to the nature of the
conditions.

On 4/22/23 2:29 PM, olcott wrote:
> On 4/22/2023 1:01 PM, Richard Damon wrote:
>> On 4/22/23 1:13 PM, olcott wrote:
>>> On 4/22/2023 11:56 AM, Richard Damon wrote:
>>>> On 4/22/23 12:45 PM, olcott wrote:
>>>>> On 4/22/2023 11:36 AM, Richard Damon wrote:
>>>>>> On 4/22/23 12:27 PM, olcott wrote:
>>>>>>> On 4/22/2023 11:12 AM, Richard Damon wrote:
>>>>>>>> On 4/22/23 11:39 AM, olcott wrote:
>>>>>>>>> On 4/22/2023 9:57 AM, Richard Damon wrote:
>>>>>>>>>> On 4/22/23 10:48 AM, olcott wrote:
>>>>>>>>>>> On 4/22/2023 9:38 AM, Richard Damon wrote:
>>>>>>>>>>>> On 4/22/23 10:28 AM, olcott wrote:
>>>>>>>>>>>>> On 4/22/2023 6:17 AM, Richard Damon wrote:
>>>>>>>>>>>>>> On 4/21/23 11:40 PM, olcott wrote:
>>>>>>>>>>>>>>> On 4/21/2023 9:45 PM, Richard Damon wrote:
>>>>>>>>>>>>>>>> On 4/21/23 9:41 PM, olcott wrote:
>>>>>>>>>>>>>>>>> On 4/21/2023 7:49 PM, Richard Damon wrote:
>>>>>>>>>>>>>>>>>> On 4/21/23 8:33 PM, olcott wrote:
>>>>>>>>>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> There exists a G such that G is logically equivalent
>>>>>>>>>>>>>>>>>>> to its own unprovability in F
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> *If we assume that there is such a G in F that means
>>>>>>>>>>>>>>>>>>> that*
>>>>>>>>>>>>>>>>>>> G is true means there is no sequence of inference
>>>>>>>>>>>>>>>>>>> steps that satisfies G in F.
>>>>>>>>>>>>>>>>>>> G is false means there is a sequence of inference
>>>>>>>>>>>>>>>>>>> steps that satisfies G in F.
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> *Thus the above G simply does not exist in F*
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> So?
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> I finally learned enough model theory to correctly link
>>>>>>>>>>>>>>>>> provability to
>>>>>>>>>>>>>>>>> truth in the conventional model theory way.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> Doesn't seem so, you don't seem to understand the
>>>>>>>>>>>>>>>> difference. You seem to confuse Truth with Knowledge.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> I finally approximated {G asserts its own unprovability
>>>>>>>>>>>>>>>>> in F}
>>>>>>>>>>>>>>>>> using conventional math symbols in their conventional way.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> Except that isn't what G is, you only think that because
>>>>>>>>>>>>>>>> you can't actually understand even the outline of
>>>>>>>>>>>>>>>> Godel's proof, so you take pieces out of context.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> G never asserts its own unprovability.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> The statement that we now have a statement that asserts
>>>>>>>>>>>>>>>> its own unprovablity, as a simplification describing a
>>>>>>>>>>>>>>>> statment DERIVED from G, and that derivation happens in
>>>>>>>>>>>>>>>> Meta-F, and is about what can be proven in F.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> Since Godel's G isn't of that form, but only can be
>>>>>>>>>>>>>>>>>> used to derive a statment IN META-F that says that G
>>>>>>>>>>>>>>>>>> is not provable in F, your argument says nothing about
>>>>>>>>>>>>>>>>>> Godel's G.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> F ⊢ GF ↔ ¬ProvF (┌GF┐).
>>>>>>>>>>>>>>>>> https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
>>>>>>>>>>>>>>>>> I have finally created a G that is equivalent to
>>>>>>>>>>>>>>>>> Panu Raatikainen's SEP article.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> So?
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> Did you read that article?
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> Also, you don't understand what those terms mean,
>>>>>>>>>>>>>>>>>> because G being true doesn't mean there is no sequence
>>>>>>>>>>>>>>>>>> of inference steps that satisfies G in F, but there is
>>>>>>>>>>>>>>>>>> no FINITE sequence of inference steps that satisfies G
>>>>>>>>>>>>>>>>>> in F.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> Because we can see that every finite or infinite
>>>>>>>>>>>>>>>>> sequence in F that
>>>>>>>>>>>>>>>>> satisfies the RHS of ↔ contradicts the LHS a powerful F
>>>>>>>>>>>>>>>>> can infer that G
>>>>>>>>>>>>>>>>> is utterly unsatisfiable even for infinite sequences in
>>>>>>>>>>>>>>>>> this more
>>>>>>>>>>>>>>>>> powerful F.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> Nope. Show the PROOF.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> You don't know HOW to do a proof, you can only do
>>>>>>>>>>>>>>>> arguement.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>>>>>>>> There exists a G in F such that G is logically equivalent
>>>>>>>>>>>>>>> to its own unprovability in F
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> A proof is any sequence of steps that shows that its
>>>>>>>>>>>>>>> conclusion is a
>>>>>>>>>>>>>>> necessary consequence of its premises.\
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Boy are you wrong.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> A proof is a FINITE sequence of steps that shows that a
>>>>>>>>>>>>>> given statement is a necessary consequence of the defined
>>>>>>>>>>>>>> system.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> "Proof" doesn't have a  "Premise", it has a system.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> The statement may have conditions in it restricting when
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>>>>>>>> There exists a G in F such that G is logically equivalent
>>>>>>>>>>>>>>> to its own unprovability in F
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> If G is true then there is no sequence of inference steps
>>>>>>>>>>>>>>> that satisfies G in F making G untrue.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> no FINITE sequence, making G UNPROVABLE, and there IS an
>>>>>>>>>>>>>> INFINITE sequence making it TRUE.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> This is possible.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> If G is false then there is a sequence of inference steps
>>>>>>>>>>>>>>> that satisfies G in F making G true.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> If G is false, then there is a finite sequence proving G,
>>>>>>>>>>>>>> which forces G to be true, thus this is a contradiction.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Because the RHS of ↔ contradicts the LHS there is no such
>>>>>>>>>>>>>>> G in F.
>>>>>>>>>>>>>>> Thus the above G simply does not exist in F.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Nope, because we can have an infinite sequence that isn't
>>>>>>>>>>>>>> finite, G can be True but not Provable.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>> If G is false and ↔ is true this makes the RHS false which
>>>>>>>>>>>>> negates the RHS making it say (G ⊢ F) which makes G true in F.
>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> Right, G can't be false, but it can be True.
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> Thus ↔ cannot be satisfied thus no such G exists in F.
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> Why do you say that?
>>>>>>>>>>
>>>>>>>>>> I don't think you know what you terms mean.
>>>>>>>>>>
>>>>>>>>>> There exists a G in F such that G is true if and only if G is
>>>>>>>>>> Unprovable.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> Logical equality
>>>>>>>>> p q p ↔ q
>>>>>>>>> T T   T // G is true if and only if G is Unprovable.
>>>>>>>>> T F   F //
>>>>>>>>> F T   F //
>>>>>>>>> F F   T // G is false if and only if G is Provable.
>>>>>>>>> https://en.wikipedia.org/wiki/Truth_table#Logical_equality
>>>>>>>>>
>>>>>>>>> Row(1) There exists a G in F such that G is true if and only if
>>>>>>>>> G is
>>>>>>>>> unprovable in F making G unsatisfied thus untrue in F.
>>>>>>>>>
>>>>>>>>> Row(4) There exists a G in F such that G is false if and only
>>>>>>>>> if G is
>>>>>>>>> provable in F making G satisfied thus true in F.
>>>>>>>>>
>>>>>>>>> If either Row(1) or Row(4) are unsatisfied then ↔ is false.
>>>>>>>>
>>>>>>>> But if neither row values can ACTUALLY EXIST, then the equality
>>>>>>>> is true.
>>>>>>>>
>>>>>>> If either Row(1) or Row(4) cannot have the same value for p and q
>>>>>>> (for whatever reason) then ↔ is unsatisfied and no such G exists
>>>>>>> in F.
>>>>>>>
>>>>>> So, you don't understand how truth tables work.
>>>>>>
>>>>>> You don't need to have all the rows with true being possible, you
>>>>>> need all the rows that are possible to be True.
>>>>>>
>>>>>
>>>>> To the best of my knowledge
>>>>> ↔ is also known as logical equivalence meaning that the LHS and the
>>>>> RHS
>>>>> must always have the same truth value or ↔ is not true.
>>>>>
>>>>
>>>> Right, and for that statement, the actual G found in F, the ONLY
>>>> values that happen is G is ALWAYS true, an Unprovable is always true.
>>>>
>>>> Thus the equivalence is always true.
>>> I don't think that is the way that it works.
>>> We must assume that the RHS is true and see how that effects the LHS
>>> We must assume that the RHS is false and see how that effects the LHS
>>> ((True(RHS) → True(LHS)) ∧ (False(RHS) → False(LHS))) ≡ (RHS ↔ LHS)
>>> False(RHS) → True(LHS) refutes (RHS ↔ LHS)
>>>
>>
>> Nope, that isn't how it works.
>>
>> Can you show me something that says that is how it works?
>>
>
> I tried and in the first search all of the articles seemed to dodge
> rather than address this point. I have always understood ↔ to mean that
> the LHS and the RHS must always have the same Boolean value.

On 4/22/2023 1:01 PM, Richard Damon wrote:
> On 4/22/23 1:13 PM, olcott wrote:
>> On 4/22/2023 11:56 AM, Richard Damon wrote:
>>> On 4/22/23 12:45 PM, olcott wrote:
>>>> On 4/22/2023 11:36 AM, Richard Damon wrote:
>>>>> On 4/22/23 12:27 PM, olcott wrote:
>>>>>> On 4/22/2023 11:12 AM, Richard Damon wrote:
>>>>>>> On 4/22/23 11:39 AM, olcott wrote:
>>>>>>>> On 4/22/2023 9:57 AM, Richard Damon wrote:
>>>>>>>>> On 4/22/23 10:48 AM, olcott wrote:
>>>>>>>>>> On 4/22/2023 9:38 AM, Richard Damon wrote:
>>>>>>>>>>> On 4/22/23 10:28 AM, olcott wrote:
>>>>>>>>>>>> On 4/22/2023 6:17 AM, Richard Damon wrote:
>>>>>>>>>>>>> On 4/21/23 11:40 PM, olcott wrote:
>>>>>>>>>>>>>> On 4/21/2023 9:45 PM, Richard Damon wrote:
>>>>>>>>>>>>>>> On 4/21/23 9:41 PM, olcott wrote:
>>>>>>>>>>>>>>>> On 4/21/2023 7:49 PM, Richard Damon wrote:
>>>>>>>>>>>>>>>>> On 4/21/23 8:33 PM, olcott wrote:
>>>>>>>>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> There exists a G such that G is logically equivalent
>>>>>>>>>>>>>>>>>> to its own unprovability in F
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> *If we assume that there is such a G in F that means
>>>>>>>>>>>>>>>>>> that*
>>>>>>>>>>>>>>>>>> G is true means there is no sequence of inference
>>>>>>>>>>>>>>>>>> steps that satisfies G in F.
>>>>>>>>>>>>>>>>>> G is false means there is a sequence of inference
>>>>>>>>>>>>>>>>>> steps that satisfies G in F.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> *Thus the above G simply does not exist in F*
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> So?
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> I finally learned enough model theory to correctly link
>>>>>>>>>>>>>>>> provability to
>>>>>>>>>>>>>>>> truth in the conventional model theory way.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Doesn't seem so, you don't seem to understand the
>>>>>>>>>>>>>>> difference. You seem to confuse Truth with Knowledge.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> I finally approximated {G asserts its own unprovability
>>>>>>>>>>>>>>>> in F}
>>>>>>>>>>>>>>>> using conventional math symbols in their conventional way.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Except that isn't what G is, you only think that because
>>>>>>>>>>>>>>> you can't actually understand even the outline of Godel's
>>>>>>>>>>>>>>> proof, so you take pieces out of context.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> G never asserts its own unprovability.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> The statement that we now have a statement that asserts
>>>>>>>>>>>>>>> its own unprovablity, as a simplification describing a
>>>>>>>>>>>>>>> statment DERIVED from G, and that derivation happens in
>>>>>>>>>>>>>>> Meta-F, and is about what can be proven in F.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> Since Godel's G isn't of that form, but only can be
>>>>>>>>>>>>>>>>> used to derive a statment IN META-F that says that G is
>>>>>>>>>>>>>>>>> not provable in F, your argument says nothing about
>>>>>>>>>>>>>>>>> Godel's G.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> F ⊢ GF ↔ ¬ProvF (┌GF┐).
>>>>>>>>>>>>>>>> https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
>>>>>>>>>>>>>>>> I have finally created a G that is equivalent to
>>>>>>>>>>>>>>>> Panu Raatikainen's SEP article.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> So?
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Did you read that article?
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> Also, you don't understand what those terms mean,
>>>>>>>>>>>>>>>>> because G being true doesn't mean there is no sequence
>>>>>>>>>>>>>>>>> of inference steps that satisfies G in F, but there is
>>>>>>>>>>>>>>>>> no FINITE sequence of inference steps that satisfies G
>>>>>>>>>>>>>>>>> in F.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> Because we can see that every finite or infinite
>>>>>>>>>>>>>>>> sequence in F that
>>>>>>>>>>>>>>>> satisfies the RHS of ↔ contradicts the LHS a powerful F
>>>>>>>>>>>>>>>> can infer that G
>>>>>>>>>>>>>>>> is utterly unsatisfiable even for infinite sequences in
>>>>>>>>>>>>>>>> this more
>>>>>>>>>>>>>>>> powerful F.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Nope. Show the PROOF.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> You don't know HOW to do a proof, you can only do arguement.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>>>>>>> There exists a G in F such that G is logically equivalent
>>>>>>>>>>>>>> to its own unprovability in F
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> A proof is any sequence of steps that shows that its
>>>>>>>>>>>>>> conclusion is a
>>>>>>>>>>>>>> necessary consequence of its premises.\
>>>>>>>>>>>>>
>>>>>>>>>>>>> Boy are you wrong.
>>>>>>>>>>>>>
>>>>>>>>>>>>> A proof is a FINITE sequence of steps that shows that a
>>>>>>>>>>>>> given statement is a necessary consequence of the defined
>>>>>>>>>>>>> system.
>>>>>>>>>>>>>
>>>>>>>>>>>>> "Proof" doesn't have a  "Premise", it has a system.
>>>>>>>>>>>>>
>>>>>>>>>>>>> The statement may have conditions in it restricting when
>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>>>>>>> There exists a G in F such that G is logically equivalent
>>>>>>>>>>>>>> to its own unprovability in F
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> If G is true then there is no sequence of inference steps
>>>>>>>>>>>>>> that satisfies G in F making G untrue.
>>>>>>>>>>>>>
>>>>>>>>>>>>> no FINITE sequence, making G UNPROVABLE, and there IS an
>>>>>>>>>>>>> INFINITE sequence making it TRUE.
>>>>>>>>>>>>>
>>>>>>>>>>>>> This is possible.
>>>>>>>>>>>>>
>>>>>>>>>>>>>> If G is false then there is a sequence of inference steps
>>>>>>>>>>>>>> that satisfies G in F making G true.
>>>>>>>>>>>>>
>>>>>>>>>>>>> If G is false, then there is a finite sequence proving G,
>>>>>>>>>>>>> which forces G to be true, thus this is a contradiction.
>>>>>>>>>>>>>
>>>>>>>>>>>>>> Because the RHS of ↔ contradicts the LHS there is no such
>>>>>>>>>>>>>> G in F.
>>>>>>>>>>>>>> Thus the above G simply does not exist in F.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>> Nope, because we can have an infinite sequence that isn't
>>>>>>>>>>>>> finite, G can be True but not Provable.
>>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> If G is false and ↔ is true this makes the RHS false which
>>>>>>>>>>>> negates the RHS making it say (G ⊢ F) which makes G true in F.
>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> Right, G can't be false, but it can be True.
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> Thus ↔ cannot be satisfied thus no such G exists in F.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> Why do you say that?
>>>>>>>>>
>>>>>>>>> I don't think you know what you terms mean.
>>>>>>>>>
>>>>>>>>> There exists a G in F such that G is true if and only if G is
>>>>>>>>> Unprovable.
>>>>>>>>>
>>>>>>>>
>>>>>>>> Logical equality
>>>>>>>> p q p ↔ q
>>>>>>>> T T   T // G is true if and only if G is Unprovable.
>>>>>>>> T F   F //
>>>>>>>> F T   F //
>>>>>>>> F F   T // G is false if and only if G is Provable.
>>>>>>>> https://en.wikipedia.org/wiki/Truth_table#Logical_equality
>>>>>>>>
>>>>>>>> Row(1) There exists a G in F such that G is true if and only if
>>>>>>>> G is
>>>>>>>> unprovable in F making G unsatisfied thus untrue in F.
>>>>>>>>
>>>>>>>> Row(4) There exists a G in F such that G is false if and only if
>>>>>>>> G is
>>>>>>>> provable in F making G satisfied thus true in F.
>>>>>>>>
>>>>>>>> If either Row(1) or Row(4) are unsatisfied then ↔ is false.
>>>>>>>
>>>>>>> But if neither row values can ACTUALLY EXIST, then the equality
>>>>>>> is true.
>>>>>>>
>>>>>> If either Row(1) or Row(4) cannot have the same value for p and q
>>>>>> (for whatever reason) then ↔ is unsatisfied and no such G exists
>>>>>> in F.
>>>>>>
>>>>> So, you don't understand how truth tables work.
>>>>>
>>>>> You don't need to have all the rows with true being possible, you
>>>>> need all the rows that are possible to be True.
>>>>>
>>>>
>>>> To the best of my knowledge
>>>> ↔ is also known as logical equivalence meaning that the LHS and the RHS
>>>> must always have the same truth value or ↔ is not true.
>>>>
>>>
>>> Right, and for that statement, the actual G found in F, the ONLY
>>> values that happen is G is ALWAYS true, an Unprovable is always true.
>>>
>>> Thus the equivalence is always true.
>> I don't think that is the way that it works.
>> We must assume that the RHS is true and see how that effects the LHS
>> We must assume that the RHS is false and see how that effects the LHS
>> ((True(RHS) → True(LHS)) ∧ (False(RHS) → False(LHS))) ≡ (RHS ↔ LHS)
>> False(RHS) → True(LHS) refutes (RHS ↔ LHS)
>>
>
> Nope, that isn't how it works.
>
> Can you show me something that says that is how it works?