Re: Full simplify problem

*To*: mathgroup at smc.vnet.net*Subject*: [mg122300] Re: Full simplify problem*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Mon, 24 Oct 2011 05:13:43 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*References*: <201110221009.GAA29840@smc.vnet.net> <201110231026.GAA10607@smc.vnet.net> <op.v3tck4nmtgfoz2@bobbys-imac.local>

In a sense, yes. But this certianly does not answer the "why" part of the OP's question and I think that was what he really wanted to know. Andrzej On 23 Oct 2011, at 18:40, DrMajorBob wrote: > Far simpler (and equivalent under the covers, I think) would be: > > E^x - E^(y + z) /. x -> y + z > > 0 > > Bobby > > On Sun, 23 Oct 2011 05:26:01 -0500, Andrzej Kozlowski <akoz at mimuw.edu.pl> wrote: > >> 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! >>> >> >> > > > -- > DrMajorBob at yahoo.com

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

**Re: Full simplify problem***From:*Andrzej Kozlowski <akoz@mimuw.edu.pl>