Variables for predicates in metagol

Question

The following program noMetagolR is given in:

http://www.doc.ic.ac.uk/~shm/Papers/metagol_gram.pdf page 33.

parse(S,G1,G2) :- parse(s(0),S,[],G1,G2).

parse(Q,X,X,G1,G2) :- abduce(acceptor(Q),G1,G2).
parse(Q,[C|X],Y,G1,G2) :- Skolem(P), abduce(delta1(Q,C,P),G1,G3), parse(P,X,Y,G3,G2).

abduce(X,G,G) :- member(X,G).
abduce(X,G,[X|G]) :- not(member(X,G)).

Skolem(s(0)). Skolem(s(1)). ...

An example query is :

parse([],[],G1), parse([0],G1,G2), parse([0,0],G2,G3), parse([1,1],G3,G4), parse([0,0,0],G4,G5), parse([0,1,1],G5,G6), parse([1,0,1],G6,G),not(parse([1],G,G)), not(parse([0,1],G,G)).

The answer substitutions should return a learnt grammar for parity.

The program is said to run in Yap. I normally use SWI-prolog. Either way, what do I do to make them understand Skolem/1 ? Presumbly this means that Skolem is a variable? I thought maybe using =.. but this does not work.

Also how many Skolem/1 facts are needed?

Answer 1

in SWI-Prolog, you can place in your source the directive

:- set_prolog_flag(allow_variable_name_as_functor,true).

see current_prolog_flag/2

example on REPL:

1 ?- set_prolog_flag(allow_variable_name_as_functor,true).
true.

2 ?- assert(X(1)).
true.

3 ?- X(Y).
Y = 1.