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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?
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