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Is there a simple way to merge equivalent conditions in Piecewise? For example, in the following we have (trivially) pw==x.

pw = Piecewise[{{x, x != 0}}, 0];

Is there some way to force this without cheating? I would call the following cheating:

Assuming[x != 0, Refine[pw]]
(* x *)

As a further example, can I get the simplification (for x real) pw2==Abs[x]?

pw2 = Piecewise[{{x, x > 0}}, -x];
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In the first case, we have

Simplify[Piecewise[{{x, x != 0}}, 0], x ∈ Reals]
(*  x  *)

In the second case, Abs[x] is a function of a complex variable and will not be treated as equivalent to Piecewise[{{x, x > 0}}, -x]. (This seems correct to me.) If one wants the standard real absolute value, one can add a domain to PiecewiseExpand:

PiecewiseExpand[Abs[x], Reals]
(*  Piecewise[{{-x, x < 0}}, x]  *)
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  • $\begingroup$ Oops -- wait -- maybe in the first case, x was supposed to be complex? $\endgroup$
    – Michael E2
    Sep 28, 2016 at 22:30
  • $\begingroup$ @ +1 for a useful partial answer. Any insight on why an irrelevant assumption (x real) helps with the first case? In the second example, I was interested in the case with x real (I think this is implicit in any expression involving an inequality). In this case, I think the expression Abs[x] is a reasonable simplification result - just as Abs[x] simplifies to x when I assumes x>0. $\endgroup$
    – mikado
    Sep 29, 2016 at 18:14
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    $\begingroup$ @mikado Not really. I thought of it immediately, actually, but I think it was for a bogus reason (now kinda forgotten -- something like the Abs[] example). Note that UnsameQ instead of Unequal works without the assumption: Simplify[Piecewise[{{x, x =!= 0}}, 0]]. -- What I've noticed about Abs[] is that M will simplify it to something else when x is real or positive, etc., but it will not go from an expression to Abs[]. I suppose there's not a strong argument why this behavior is preferably, but I almost never want Abs[x] when x is real (differentiating it is a pain). $\endgroup$
    – Michael E2
    Sep 29, 2016 at 18:33

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