Re: Full simplify problem

*To*: mathgroup at smc.vnet.net*Subject*: [mg122333] Re: Full simplify problem*From*: "A. Lapraitis" <ffcitatos at gmail.com>*Date*: Tue, 25 Oct 2011 06:18:59 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*References*: <D3C63570-91CA-489E-B8B7-661844B401BC@mimuw.edu.pl>

First of all, thanks to all the people answering! The problem is more complicated, indeed, this is just the minimal working example. However, I still do not understand Mathematica's behaviour. 1. I thought that the whole point of Count[#, x, Infinity]& was to distinguish between Exp[x] and Exp[y+z] in favour of Exp[x]. However, the same is true for the LeafCount function (it gives 3 and 5 respectively). Why is not a good complexity function in this case? 2. I was trying a different workaround and have noticed a dependence of the result on the names of the variables: In[13]:= Clear["Global`*"] In[14]:= Assuming[x == d && d == y + z, FullSimplify[E^x - E^(y + z)]] Out[14]= E^x - E^(y + z) In[15]:= Assuming[a == d && d == b + c, FullSimplify[E^a - E^(b + c)]] Out[15]= 0 Can anyone explain this?

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

**Re: Full simplify problem***From:*Daniel Lichtblau <danl@wolfram.com>