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  • $\begingroup$ Does this work for you: f[expr_ /; ! FreeQ[expr, _F] && ! FreeQ[expr, _G]] := 0; f[expr_] := expr ? $\endgroup$ Commented Nov 13, 2019 at 12:15
  • $\begingroup$ Thanks for your answer. Unfortunately it does not work because I specifically need F and G to also have the opposite arguments as stated F[X] and G[-X]. $\endgroup$ Commented Nov 13, 2019 at 12:26
  • $\begingroup$ Ah, maybe this one: f[expr_ /; ! FreeQ[expr, F[x_]] && ! FreeQ[expr, G[-x_]]] := 0; f[expr_] := expr. tested with f[F[Sin[x]]*G[Sin[-x]]] is 0 because MMA replaces G[Sin[-x]] with G[-Sin[x]]. $\endgroup$ Commented Nov 13, 2019 at 13:05
  • $\begingroup$ Yes, this works at first glance but I am not really sure it distinguishes the x argument correctly that it is the same in both F and G. Try this for example f[F[X] G[-Z]] this is evaluated to zero when it shouldn't since they do not have the same argument. $\endgroup$ Commented Nov 13, 2019 at 15:40