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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:= Clear["Global`*"] In:= Assuming[x == d && d == y + z, FullSimplify[E^x - E^(y + z)]] Out= E^x - E^(y + z) In:= Assuming[a == d && d == b + c, FullSimplify[E^a - E^(b + c)]] Out= 0 Can anyone explain this?