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    <p>Haskell will perform what is known as lazy evaluation, meaning it will only work things out as it needs them from left to right (generally). So taking your example of <code>mystery [1, 2]</code>, Haskell will do the following:</p> <pre><code>sets ++ (map (x:) sets) </code></pre> <p>Which evaluates to:</p> <pre><code>mystery (2:[]) ++ (map (1:) sets) </code></pre> <p>At this point, we're calling <code>mystery (2:[])</code></p> <pre><code>mystery ([]) ++ (map (2:) sets) ++ (map (1:) sets) </code></pre> <p><code>mystery ([])</code> will return an empty list of lists</p> <pre><code>[[]] ++ (map (2:) sets) ++ (map (1:) sets) [[]] ++ (map (2:) mystery []) ++ (map (1:) sets) </code></pre> <p>So now Haskell will try to apply the function <code>(2:)</code> on a list containing an empty list</p> <pre><code>[[]] ++ (2:[[]]) ++ (map (1:) sets) [[]] ++ [[2]] ++ (map (1:) sets) [[], [2]] ++ (map (1:) sets) </code></pre> <p>This is where things get a little more confusing.</p> <pre><code>[[], [2]] ++ (map (1:) mystery (2:[])) </code></pre> <p>That last <code>sets</code> will evaluate <code>mystery (2:[])</code></p> <pre><code>[[], [2]] ++ (map (1:) (sets ++ (map (2:) sets))) [[], [2]] ++ (map (1:) (mystery [] ++ (map (2:) sets)) [[], [2]] ++ (map (1:) ([[]] ++ (map (2:) mystery [])) [[], [2]] ++ (map (1:) ([[]] ++ (2:[[]])) [[], [2]] ++ (map (1:) ([[]] ++ [[2]]) </code></pre> <p>Now <code>(1:)</code> will be applied to a list which contains an empty list, and a list containing the 2:</p> <pre><code>[[], [2]] ++ (map (1:) ++ [[], [2]]) [[], [2]] ++ [[1], [1, 2]] [[], [2], [1], [1, 2]] </code></pre> <p>The real meat of the operation is in those last two sections. Haskell creates a list like <code>[[], [2]]</code> and then appends one to the head of each list to form <code>[[1], [1, 2]]</code>.</p>
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    1. COWow! Thank you for such a thorough response! Much appreciated.
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    2. COYou're very welcome!
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