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Proposition: If a function is injective, its inverse is also a function:
forall A, B, f: A ->injective B: : f-1: B -> A.
Proof: Take arbitrary f: A ->injective B. We have to show f-1: B -> A.
We have f-1 subset B x A. Thus it remains to be shown
(forall x,y0, y1: (<x, y0> in f-1 /\ <x, y1> in f-1) => y0=y1).Take arbitrary x, y0, and y1 and assume
(1) <x, y0> in f-1 /\ <x, y1> in f-1.We have to show (2) y0=y1. From (1) and the definition of inverse, we know
(3) <y0, x> in f /\ <y1, x> in f,i.e., f(y0) = x and f(y1) = x. Since f is injective, we thus know (2).