#### Answer

Prolog will conclude that Carol is her own sibling. To solve this problem, the rule needs to include the fact that X cannot be equal to Y, which in Prolog is written X \ = Y. Thus an improved version of the rule would be
sibling (X, Y) :-X \= Y, parent(Z, X), parent(Z, Y).
which says that X is Y’s sibling if X and Y are not equal and have a common parent. The following version would insist that X and Y are siblings only if they have both parents in common:
sibling (X, Y) :− X \= Y, Z \= W
parent (Z, X), parent (Z, Y),
parent (W, X), parent (W, Y).