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    <p>I think the problem is the 'implicit cut' in (->)/2. Try</p> <pre><code>test_arith(Passes, Failures) :- findall(P-F, (test_arith, P=1,F=0 ; P=0,F=1), Scores), summarize_scores(Scores, 0, 0, Passes, Failures). </code></pre> <p>and you'll get</p> <pre><code>?- test_arith(P,F). P = 2, F = 1. </code></pre> <p><em>edit</em></p> <p>OT, but I like when I can make the logic more compact, of course with a little help from the library. Here an equivalent definition:</p> <pre><code>test_arith(Passes, Failures) :- findall(R, (test_arith, R=1-0 ; R=0-1), Scores), aggregate(r(sum(A),sum(B)), member(A-B, Scores), r(Passes, Failures)). </code></pre> <p>and then, why to build a list to be immediately scanned?</p> <pre><code>test_arith(Passes, Failures) :- aggregate(r(sum(A),sum(B)), (test_arith, A=1,B=0 ; A=0,B=1), r(Passes, Failures)). </code></pre> <p><em>edit</em> the above code is incorrect, being unable to count failures. I was fooled by the fact that it seemed to work with the specific test case.</p> <p>With the help of @false, here is reify_call/3, a building block that could solve OP' problem (tested in SWI-Prolog, where <a href="http://www.swi-prolog.org/pldoc/doc_for?object=clause/2" rel="nofollow">clause</a>/2 is arguably extended with respect to ISO compatibility, given @false comment to question):</p> <pre><code>test_arith(Passes, Failures) :- aggregate(r(sum(T),sum(F)), reify_call(test_arith, T, F), r(Passes, Failures)). :- meta_predicate reify_call(0, -, -). reify_call(Pred, True, False) :- clause(Pred, Cl), (call(Cl) -&gt; True = 1, False = 0 ; True = 0, False = 1). </code></pre>
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    1. POProlog: How do I count the number of passes and failures for a predicate?
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    1. COYou are incorrectly determining the number of failures: You will have always exactly one failure...
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