Challenge 2: Extending solve/1 With Disjunction and Negation — Possible Solution
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meta_demo.pl:

solve(true) :- !.
solve((A, B)) :- !, solve(A), solve(B).
solve((A ; B)) :- !, (solve(A) ; solve(B)).
solve(\+ A) :- !, \+ solve(A).
solve(A) :- clause(A, Body), solve(Body).

likes(tom, pizza).
likes(tom, sushi).

Queries:

    ?- solve((likes(tom, pizza) ; likes(tom, tacos))).
    true.

    ?- solve(\+ likes(tom, tacos)).
    true.

Explanation:
The first query matches the new disjunction clause: solve((A ; B))
tries solve(likes(tom, pizza)) first, which succeeds directly via
clause/2 (a matching fact), so the whole disjunction succeeds without
ever needing to try the tacos branch. The second query matches the
negation clause: solve(\+ A) calls \+ solve(likes(tom, tacos))
internally -- solve(likes(tom, tacos)) itself fails, since no such
fact or rule exists for clause/2 to find, so \+ around that failed
call succeeds, and solve/1 reports true. Both new clauses work by
directly reusing Prolog's own real ; and \+ on the result of a
recursive solve/1 call, exactly as the chapter describes -- no
separate disjunction- or negation-solving logic was written from
scratch.

WHY THIS WORKS AS AN ANSWER
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This adds the chapter's own disjunction and negation clauses
unmodified, then exercises both with a fresh likes/2 example (not
reusing the chapter's own parent/grandparent facts), confirming each
new clause correctly delegates to Prolog's real ; and \+ rather than
reimplementing their logic.