$Mx =_{\beta\eta} Nx$ implies $M =_{\beta\eta} N$

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In the context of $\lambda$-calculus, I was thinking about whether or not $$Mx =_{\beta\eta} Nx \implies M =_{\beta\eta} N$$ if $x\notin FV(M)\cup FV(N)$. I have been around this issue for quite some time now, but didn't reach anywhere useful.

Is it the case or not?