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).
![](https://gradesaver.s3.amazonaws.com/uploads/solution/dd8ab0e5-8905-409d-814a-8625a7b7fb02/result_image/1560969983.png?X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Credential=AKIAJVAXHCSURVZEX5QQ%2F20240727%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Date=20240727T010947Z&X-Amz-Expires=900&X-Amz-SignedHeaders=host&X-Amz-Signature=8ab34f9e1c6e56aff2d1dc7df9a67ed778cf3339dda62a93f7dab4d4d8bf4d60)
Work Step by Step
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).
![](https://gradesaver.s3.amazonaws.com/uploads/solution/dd8ab0e5-8905-409d-814a-8625a7b7fb02/steps_image/small_1560969983.png?X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Credential=AKIAJVAXHCSURVZEX5QQ%2F20240727%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Date=20240727T010947Z&X-Amz-Expires=900&X-Amz-SignedHeaders=host&X-Amz-Signature=b68c31a2bbb3c36a79e3b13600e56dc5be8cf22c790bd5aeb45aa175843f6392)