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Re: Full simplify problem

On 10/25/2011 05:18 AM, A. Lapraitis wrote:
> 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?

Daniel Lichtblau
Wolfram Research

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