Nuprl Lemma : bijection_restriction 4,23

k:f:(kk).
0<k  Bij(kkf f(k-1) = k-1    {f  (k-1)(k-1) & Bij((k-1); (k-1); f)} 
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DefinitionsProp, {T}, S  T, S  T, x:AB(x), i  j < k, P & Q, t  T, , AB, A, False, P  Q, Bij(ABf), {i..j}, T, True, Dec(P), P  Q, Inj(ABf), Surj(ABf), x:AB(x)
Lemmasle wf, decidable int equal, nat wf, int seg wf, biject wf

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