I have the following issue with FourierTransform, Fourier transform doesn't seem to know it is a linear operator.

aa = FourierTransform[s1[t],t,omega];
bb = FourierTransform[s1'[t],t,omega];

cc = FourierTransform[s1[t] + s1'[t],t,omega];

Expand[aa+bb] == Expand[cc]

This yields false (more precisely, it doesn't yield true). I would expect the Fourier transform to know it is a linear operator. Do I have to do something like the following?

fourierTransformSubs = 
FourierTransform[a1_ f_[t_],t_,omega_] :>  a1 FourierTransform[f[t],t,omega],
FourierTransform[f_[t_]+g_[t_] ,t_,omega_] :> FourierTransform[f[t],t,omega] + 
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    $\begingroup$ Welcome to Mathematica.SE! I hope you will become a regular contributor. To get started, 1) take the introductory tour now, 2) when you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge, 3) remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign, and 4) give help too, by answering questions in your areas of expertise. $\endgroup$ – bbgodfrey Sep 26 '15 at 1:18
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    $\begingroup$ Are you asking how to show the identity with s[t] defined or undefined? $\endgroup$ – bbgodfrey Sep 26 '15 at 1:36
  • $\begingroup$ s[t] is undefined. I want to do things symbolically for now. $\endgroup$ – fred Sep 26 '15 at 1:49
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    $\begingroup$ Well, it's an interesting question what should happen for the symbolic case. It's possible that a sum has a Fourier transform but its terms don't. $\endgroup$ – John Doty Sep 26 '15 at 2:59
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    $\begingroup$ Related: mathematica.stackexchange.com/a/71393/1871 $\endgroup$ – xzczd Sep 26 '15 at 8:46