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Re: Full simplify problem

  • To: mathgroup at smc.vnet.net
  • Subject: [mg122289] Re: Full simplify problem
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Sun, 23 Oct 2011 06:26:01 -0400 (EDT)
  • Delivered-to: l-mathgroup@mail-archive0.wolfram.com
  • References: <201110221009.GAA29840@smc.vnet.net>

Because FullSimplify would have to increase the default complexity to 
obtain the cancellation. But this works:

Assuming[x == y + z,
 FullSimplify[E^x - E^(y + z),
  ComplexityFunction -> (Count[#, x, {1, Infinity}] &)]]

0

Note that the reason why this works is that:

Assuming[x == y + z,
 FullSimplify[x, ComplexityFunction -> (Count[#, x, {1, Infinity}] &)]]

 y+z


Andrzej Kozlowski


On 22 Oct 2011, at 12:09, A. Lapraitis wrote:

> Hello,
>
> Could anyone explain why the following does not give zero?
>
> In[72]:= Assuming[
> x == y + z,
> FullSimplify[
>  E^x - E^(y + z)
>  ]
> ]
>
> Out[72]= E^x - E^(y + z)
>
> Thanks!
>




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