Nuprl Lemma : fpf-sub_weakening 11,40

A:Type, B:(AType), eq:EqDecider(A), f,g:fpf(Aa.B(a)).
(f = g fpf-sub(Aa.B(a); eqfg
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Definitionsx:AB(x), x(s), P  Q, t  T, prop{i:l}, xt(x), fpf-sub(Aa.B(a); eqfg), A c B
Lemmasfpf-sub wf, fpf wf, deq wf, fpf-ap wf, assert wf, fpf-dom wf, fpf-trivial-subtype-top

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