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Assuming $f^{-1}$ exists (i.e., that $f$ is injective), then $$f_k^{-1}(y) = f^{-1}(y \oplus k) \oplus k.$$ To see why consider an $x \in \{0,1\}^4$ so that $f_k(x) = y$. Then you have that $$y \oplus k = f (x \oplus k).$$ Now by applying $f^{-1}$ on both sides of that equation you have $$f^{-1}(y \oplus k) = x \oplus k.$$ And finally by XOR'ing $k$ on both ...



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