Nuprl Lemma : strict-fun-connected-step 11,40

T:Type, f:(TT), x:T. ((f(x) = x))  f(x) = f+(x
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Definitionsy = f+(x), P & Q, x:A  B(x), f(a), Void, Type, x:AB(x), x:AB(x), s = t, P  Q, False, A, y is f*(x), [car / cdr], [], A List, last(L), <ab>, , , Unit, (xL.P(x)), xLP(x), |r|, x f y, A c B, a < b, |g|, a <p b, a  b, |p|, a ~ b, b | a, x:AB(x), x,y:A//B(x;y), b, Atom, Dec(P), P  Q, left + right, s ~ t, , SQType(T), {T}, tl(l), i j, i <z j, hd(l), l[i], n+m, #$n, a < b, {i..j}, {x:AB(x)} , , i  j < k, A  B, , type List, y=f*(x) via L, t  T
Lemmasfun-path wf, guard wf, decidable int equal, le wf, int seg wf, false wf, not wf, fun-connected wf, strict-fun-connected wf

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