Matching in SICStus prolog - prolog

Question

This is my code, which is for Satchmo theorem proving. It does some unification.
:- op(700, xfx, ==>).
:- op(400, yfx, &).
:- op(400, yfx, or).
fact([a, 9]).
fact([b, 9]).
rule([a, X] & [b, X] ==> [c, X]). %% horn bit
rule([c, X] ==> [r, X] or [s, X]). %% non horn bit
rule([r, X] ==> [t, X]).
rule([s, X] ==> [t, X]).
horn(A & B) :-
!,
horn(A),
horn(B).
horn(A or B) :-
!,
(horn(A); horn(B)).
horn(P) :-
fact(P).
horn(P) :-
temp(P).
horn(P) :-
rule(SUBGOALS ==> P),
\+ P = (_A or _B),
horn(SUBGOALS).
satchmo(P) :-
retractall(temp(_)),
prove(P).
prove(P) :-
horn(P).
prove(P) :-
rule(LHS ==> (A or B)),
horn(LHS),
\+ horn(A or B),
cprove(A ==> P),
cprove(B ==> P).
cprove(A ==> P) :-
try(A),
(prove(P) ->
untry(A);
(untry(A), fail)).
try(A & B) :-
!,
try(A),
try(B).
try(A) :-
assert(temp(A)).
untry(A & B) :-
!,
untry(A),
untry(B).
untry(A) :-
retract(temp(A)).
To understand how is it working, we can do that by ?- spy([satchmo]).
1- if the given query is fact as:
?- satchmo([a, 9]).
yes.
or
?- satchmo([b, 9]).
yes.
the program will prove it as it is a fact.
2- If the query in the horn bit as:
?- satchmo([c, 9]).
yes.
the program will prove it as it is the horn rule.
3- If the query in the non-horn bit as:
?- satchmo([t, 9]).
yes.
It'll be proved as well.
This is working perfectly. But when I have tried to change it a bit. Instead of unification I need to do another kind of matching which can prove the following:
if I have:
rule[living, X] ==> [mortal, X].
[man, socrates].
I would like to prove:
?- satchmo([mortal, socrates]).
yes.
To do this I have amend my code a bit, so instead of having:
horn(P):-
fact(P).
I put something very similar:
horn(P):-
match(P, P0),
fact(P0).
and I have defined match as:
match(X, Y):-
X=Y.
I know that I have not do anything by this move, but I am thinking to amend the definition of match a bit to be able to prove what I need.
I am stuck somewhere here, look at my current code.
:- op(700, xfx, ==>).
:- op(400, yfx, &).
:- op(400, yfx, or).
:- op(400, yfx, <<<).
fact([a, 9]).
fact([b, 9]).
rule([a, X] & [b, X] ==> [c, X]). %% horn bit
rule([c, X] ==> [r, X] or [s, X]). %% non horn bit
rule([r, X] ==> [t, X]).
rule([s, X] ==> [t, X]).
man <<< human.
human <<< animal.
animal <<< living.
subset(X, X).
subset(X, Y) :-
X <<< Y.
subset(X, Z) :-
X <<< Y,
subset(Y, Z).
horn(A & B) :-
!,
horn(A),
horn(B).
horn(A or B) :-
!,
(horn(A); horn(B)).
horn(P) :-
fact(P).
horn(P) :-
temp(P).
horn(P) :-
rule(SUBGOALS ==> P),
\+ P = (_A or _B),
horn(SUBGOALS).
satchmo(P) :-
retractall(temp(_)),
prove(P).
prove(P) :-
horn(P).
prove(P) :-
rule(LHS ==> (A or B)),
horn(LHS),
\+ horn(A or B),
cprove(A ==> P),
cprove(B ==> P).
cprove(A ==> P) :-
try(A),
(prove(P) ->
untry(A);
(untry(A), fail)).
try(A & B) :-
!,
try(A),
try(B).
try(A) :-
assert(temp(A)).
untry(A & B) :-
!,
untry(A),
untry(B).
untry(A) :-
retract(temp(A)).
Here, we can test the subset:
?- subset(human, man).
yes.
The problem here, is I do not know how to achieve my goal by proving:
?- satchmo([mortal, socrates]).
yes.
from:
[living, X] ==> [mortal, X].
[man, socrates].
Could it be done by changing the definition of match? If not, is there any other method to do it?

Answer 1

Ok, done,
The solution of this problem is simply by changing the definition of match, so it will be as follows:
match(P, P0) :-
P = [X, _],
P0 = [Y, _],
subset(Y, X).
subset(X, X).
subset(X, Y) :-
X <<< Y.
subset(X, Z) :-
X <<< Y,
subset(Y, Z).
and when I have this facts:
fact([man, socrates]).
man <<< human.
human <<< animal.
animal <<< living.
and this rule:
rule([living, X] ==> [mortal, X]).
I can prove:
?- satchmo([mortal, socrates]).