Finished Prolog Interpreter! yehhhh ... could be some tweaks
as i get more advanced programs in Prolog

will go LIVE soon few cosmetic changes

Now it DISPLAYS the path at the end

go FROM TO TO
:-
rd FROM TO

go FROM TO [ p FROM PTH ]
:-
rd FROM MID
go MID TO PTH

query :- go 1 7 PATH
-----------
---------- BEGIN 2 ---------
go 1 7 PATH
2>>> F = 1
2>>> T = 7
2>> PATH = P

RULE 1
3>> FROM = 1
3>> TO = 7

-----------
go FROM TO TO :- rd FROM TO
-----------
---------- BEGIN 4 ---------
rd FROM TO
3> FROM = 1
3> TO = 7
NOT FOUND

****** 3 ******
F = 1 T = 7 PATH = P FROM = 1 TO = 7
****************

RULE 2
5>> FROM = 1
5>> TO = 7
5>>> P = [ p FROM PTH ]

-----------
go FROM TO [ p FROM PTH ] :- rd FROM MID go MID TO PTH
-----------
---------- BEGIN 6 ---------
rd FROM MID
5> FROM = 1
6>> MID = 3
---------- BEGIN 7 ---------
go MID TO PTH
6> MID = 3
6> TO = 7
7>>> F = 3
7>>> T = 7
7>> PTH = FROM-->1
PTH-->PTH
[ p 1 PTH ]

RULE 1
8>> FROM = 3
8>> TO = 7

-----------
go FROM TO TO :- rd FROM TO
-----------
---------- BEGIN 9 ---------
rd FROM TO
8> FROM = 3
8> TO = 7
NOT FOUND

****** 8 ******
F = 3 T = 7 PTH = [ p 1 PTH ] FROM = 3 TO = 7
****************

RULE 2
10>> FROM = 3
10>> TO = 7
10>>> P = [ p FROM PTH ]

-----------
go FROM TO [ p FROM PTH ] :- rd FROM MID go MID TO PTH
-----------
---------- BEGIN 11 ---------
rd FROM MID
10> FROM = 3
11>> MID = 5
---------- BEGIN 12 ---------
go MID TO PTH
11> MID = 5
11> TO = 7
11> PTH = [ p 1 PTH ]
12>>> F = 5
12>>> T = 7
12>>> P = [ p 1 PTH ]

RULE 1
13>> FROM = 5
13>> TO = 7

-----------
go FROM TO TO :- rd FROM TO
-----------
---------- BEGIN 14 ---------
rd FROM TO
13> FROM = 5
13> TO = 7

****** 13 ******
F = 5 T = 7 P = [ p 1 PTH ] FROM = 5 TO = 7
****************

****** 10 ******
F = 3 T = 7 PTH = [ p 1 PTH ] FROM = 3 TO = 7 P = [ p FROM PTH ]
****************

****** 5 ******
F = 1 T = 7 PATH = P FROM = 1 TO = 7 P = [ p FROM PTH ]
****************

YES

PATH = [ p 5 [ p 1 PTH ] ]