Re: Full simplify problem

*To*: mathgroup at smc.vnet.net*Subject*: [mg122310] Re: Full simplify problem*From*: DrMajorBob <btreat1 at austin.rr.com>*Date*: Mon, 24 Oct 2011 05:15:31 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*References*: <201110221009.GAA29840@smc.vnet.net>*Reply-to*: drmajorbob at yahoo.com

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>