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Re: Full simplify problem


Far simpler (and equivalent under the covers, I think) would be:

E^x - E^(y + z) /. x -> y + z

0

Bobby

On Sun, 23 Oct 2011 05:26:01 -0500, Andrzej Kozlowski <akoz at mimuw.edu.pl>  
wrote:

> Because FullSimplify would have to increase the default complexity to
> obtain the cancellation. But this works:
>
> Assuming[x == y + z,
>  FullSimplify[E^x - E^(y + z),
>   ComplexityFunction -> (Count[#, x, {1, Infinity}] &)]]
>
> 0
>
> Note that the reason why this works is that:
>
> Assuming[x == y + z,
>  FullSimplify[x, ComplexityFunction -> (Count[#, x, {1, Infinity}] &)]]
>
>  y+z
>
>
> Andrzej Kozlowski
>
>
> On 22 Oct 2011, at 12:09, A. Lapraitis wrote:
>
>> Hello,
>>
>> Could anyone explain why the following does not give zero?
>>
>> In[72]:= Assuming[
>> x == y + z,
>> FullSimplify[
>>  E^x - E^(y + z)
>>  ]
>> ]
>>
>> Out[72]= E^x - E^(y + z)
>>
>> Thanks!
>>
>
>


-- 
DrMajorBob at yahoo.com



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