Nuprl - Proof: bij imp exists inv 1 1 1 1 1 1

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At: bij imp exists inv 1 1 1 1 1 1

1. A : Type
2. B : Type
3. f : A

B
4.

a1,a2:A. f(a1) = f(a2)

a1 = a2
5. g : B

A
6.

b:B. f(g(b)) = b

g o f = Id

By:	(Unfolds [`identity`;`compose`] 0 THEN Ext) THEN (Reduce 0)

Generated subgoal:

1 7. x : A
g(f(x)) = x
2 steps

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