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Re: Full simplify problem


Much easier is

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

0

...unless your real problem is more complicated?

Bobby

On Sun, 23 Oct 2011 05:24:44 -0500, dimitris <dimmechan at yahoo.com> wrote:

> On Oct 22, 1:18 pm, "A. Lapraitis" <ffcita... at gmail.com> 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!
>
> Actually private communication with Andrzej Kozlowski   made me
> understand that the approach in my second message was incorrect.
> For example each of the following evaluate to True
>
> In[116]:= Assuming[x === y + z, FullSimplify[E^x - E^(y + z) ===
>  0]]
>
> During evaluation of In[116]:= $Assumptions::fas: Warning: One or more
> assumptions evaluated to False. >>
>
> Out[116]= True
>
> In[119]:= Assuming[x === y + z, FullSimplify[E^x - E^(y + z) < 0]]
>
> During evaluation of In[119]:= $Assumptions::fas: Warning: One or more
> assumptions evaluated to False. >>
>
> Out[119]= True
>
> In[120]:= Assuming[x === y + z, FullSimplify[E^x - E^(y + z) > 0]]
>
> During evaluation of In[120]:= $Assumptions::fas: Warning: One or more
> assumptions evaluated to False. >>
>
> Out[120]= True
>
> Assuming False every predicate becomes True.
>
> Of course the approach in the first message is correct.
>
> Dimitris
>


-- 
DrMajorBob at yahoo.com



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