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I have encountered some problems when using Simplify with expressions containing a square root and isolated a following test case

Simplify[x - #[y], x == #[y]] & /@ {Sin, Sqrt, #^s &, #^(3/2) &}

with a result

{0, x - Sqrt[y], 0, x - y^(3/2)}
  1. Why is x-Sqrt[y] not simplified to 0 as in the case of x-Sin[y]?
  2. What is the logic behind simplifying x-y^s to 0, but keeping x-y^(3/2)?
  3. Why do all the above differences vanish when we replace - with ==? In such case

    Simplify[x == #[y], x == #[y]] & /@ {Sin, Sqrt, #^s &, #^(3/2) &}

    results in

    {True, True, True, True}
share|improve this question
FWIW, Sqrt[y]'s FullForm is really Power[y, Rational[1/2]], just as y^(3/2)'s Fullform is Power[y, Rational[3/2]], so it seems likely that the same cause (fractional powers?) is leading to the observed output in the two cases. – Aky Dec 11 '13 at 7:45
Furthermore: Simplify[x - y^(3/2) == 0, x == y^(3/2)] ==> True but Simplify[PossibleZeroQ[x - y^(3/2)], x == y^(3/2)] ==> False. – István Zachar Dec 11 '13 at 8:18
BTW, was it confirmed by WRI or TechSupport that this is a bug? Until it is approved, I remove the tag bugs, please only use it if there is an authentic comment from WRI or reasonable community-backing that a behaviour is indeed a bug. – István Zachar Dec 11 '13 at 8:19
I have not reported this behavior via official channels yet. @Aky You are of course right with the FullForm of Sqrt, it is no different than the case x == y^(3/2). But still, why x == y^s behaves in a different way? – pwl Dec 11 '13 at 10:19

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