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! >

**References**:**Full simplify problem***From:*"A. Lapraitis" <ffcitatos@gmail.com>