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Re: Full simplify problem


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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