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Reuse results #15

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japgolly opened this issue Nov 13, 2015 · 0 comments
Open

Reuse results #15

japgolly opened this issue Nov 13, 2015 · 0 comments

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@japgolly
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In (A ∧ B) ∨ (¬A ∧ C), A shouldn't be evaluated twice.

Maybe it's already smart like that; can't remember. Check.

Also it should be lazy such that C isn't evaluated when (A ∧ B) is true. Maybe that's already the case too (?).
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@japgolly