Nuprl - Proof: bij imp exists inv2

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Bijections have inverses.

A,B:Type, f:(A

B). Bij(A; B; f)

(

g:(B

A). InvFuns(A;B;f;g))

By:	Repeat Def of InvFuns(A;B;f;g) \| Bij(A; B; f) \| Inj(A; B; f) \| Surj(A; B; f)

Generated subgoal:

1 1. A : Type
2. B : Type
3. f : AB
4. a1,a2:A. f(a1) = f(a2) a1 = a2
5. b:B. a:A. f(a) = b
g:(BA). (x:A. g(f(x)) = x) & (y:B. f(g(y)) = y)
7 steps

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