Re: Full simplify problem

*To*: mathgroup at smc.vnet.net*Subject*: [mg122295] Re: Full simplify problem*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Mon, 24 Oct 2011 05:12:49 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*References*: <D3C63570-91CA-489E-B8B7-661844B401BC@mimuw.edu.pl> <C9A65308-E4DD-4DD6-967D-FFDF15F6E7DC@mimuw.edu.pl> <1319314368.91650.YahooMailNeo@web43134.mail.sp1.yahoo.com>

This is an interesting example! Here is the explanation and a puzzle: In[28]:= Assuming[False, Simplify[1 == 2]] During evaluation of In[28]:= $Assumptions::fas: Warning: One or more assumptions evaluated to False. >> Out[28]= True Do you see the relation ;-) Andrzej On 22 Oct 2011, at 22:12, dimitris anagnostou wrote: > With your help I understood why Assuming[x ==y + z, FullSimplify[E^x - E^(y + z)]] > does not work here. > What is normal for me, is not normal for such advanced function like FullSimplify! > > What about the following approach? > > In[83]:= Assuming[x === y + z, FullSimplify[E^x - E^(y + z)]] > > During evaluation of In[83]:= $Assumptions::fas:Warning: One or more assumptions evaluated to False. >> > > Out[83]= 0 > > Apart from the warning message does the requested job. > > I do consider however the solution with ComplexityFunction much more clever and correct of course! > > Dimitris > > > From: Andrzej Kozlowski <akoz at mimuw.edu.pl> > To: dimitris anagnostou <dimmechan at yahoo.com> > Sent: Saturday, October 22, 2011 10:45 PM > Subject: Fwd: Full simplify problem > > Hello Dimitris > > I also sent a response to that message. It is quite similar to yours. > > Best regards > > Andrzej > > Begin forwarded message: > >> From: Andrzej Kozlowski <akoz at mimuw.edu.pl> >> Date: 22 October 2011 17:02:05 CEST >> To: A. Lapraitis <ffcitatos at gmail.com> >> Cc: mathgroup at smc.vnet.net >> Subject: Re: Full simplify problem >> >> 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! >>> >> > > >