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copied!<p>It's a bit unclear to me what you want to achieve with your predicate - should it print one student at a time and abort if the user accepts, or should it simply print out all sad students in the list? These would be two slightly different predicates.</p> <p>As to why you're getting duplicates, I guess something in your <code>sad</code> predicate leads to backtracking, meaning it can be evaluated multiple times - use the <code>trace</code> functionality of the prolog interpreter to find out where the backtracking happens. While the best thing to do would be fixing the <code>sad</code> predicate to not backtrack, a workaround would be using a cut after the <code>sad</code> which prevents the backtracking:</p> <pre><code>calculate_sad_moods([]):-!. calculate_sad_moods([X2|Rest2]) :- (sad(X2), !, write(`One of our sad students is: `), write(X2), nl, calculate_sad_moods(Rest2)); calculate_sad_moods(Rest2). </code></pre> <p>The cut means that we need to recurse into <code>calculate_sad_moods</code> after writing the output as we can't reach the <code>or</code> part of the clause anymore (it's been cut away too).</p> <p>If you require the behaviour that the predicate evaluates to true after the first sad student found you could achieve this by using another <code>;</code> before the backtracking in the sadness case:</p> <pre><code>calculate_sad_moods([X2|Rest2]) :- (sad(X2), !, write(`One of our sad students is: `), write(X2), (nl; calculate_sad_moods(Rest2))); %only recurse if user says 'no' calculate_sad_moods(Rest2). </code></pre> <p>As a final note, the predicate name is very misleading as you don't actually calculate anything; maybe <code>print_sad_students</code> would be a more fitting name.</p>

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