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copied!<p>Given that you've got <code>female/1</code> & <code>male/1</code> predicates the predicate becomes quite simple.</p> <pre><code>marriedcouple(X,Y) :- parent( (X,Z), bornin(_) ), parent( (Y,Z), bornin(_) ), female(X), male(Y). </code></pre> <p>However, if you want to see if X and Y are not the same use the <code>(\==)/2</code> operator for "not identical" or <code>(\=)/2</code> for "not unifiable".</p> <hr/> <blockquote> <p>Pradeep, based on your comment below, here is a more complete solution.</p> </blockquote> <p>In order to prevent the same answer coming back twice there's a number of choices. We can build a list of solutions and only add a newly found solution if it isn't already in the list. Or use an approach that incorporates state using the <code>assert/1</code> predicate.</p> <p>I've chosen the latter.</p> <pre><code>?- solve. solve :- marriedcouple(Dad, Mum), not( found( marriedcouple(Dad, Mum) ) ), assert( found( marriedcouple(Dad, Mum) ) ), write( [Dad, Mum] ), nl, fail. marriedcouple(Dad, Mum) :- parent(Dad, Child), parent(Mum, Child), male(Dad), female(Mum). male(aaron). male(adam). female(betty). female(eve). parent(aaron, callum). parent(aaron, david). parent(adam, abel). parent(adam, cain). parent(betty, callum). parent(betty, david). parent(eve, abel). parent(eve, cain). </code></pre> <p>When I run this I get the following:</p> <pre><code>[aaron,betty]; [adam,eve]; No. </code></pre> <p>Be careful using <code>assert/1</code> predicates as you may introduce unwanted side-effects into your programs. You may need to do appropriate <code>retract/1</code> calls too.</p>

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