Simplify NaNMath.pow substitution
#689
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Does this not lose the |
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Initially I assumed it would, but based on the tests I ran it didn't, which is why I opted for removing this special branch completely. My hypothesis was that it is already covered by the existing rewriting/simplification performed for |
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The |
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Line 149 was wrong, but it seems it was not caught by the tests since this simplification is applied already when the expression is constructed. Which made me wonder whether we could remove this branch completely from the
NaNMath.powsubstitution.