Nuprl Lemma : fpf-cap-compatible
∀[X:Type]. ∀[eq:EqDecider(X)]. ∀[f,g:x:X fp-> Type]. ∀[x:X].
(f(x)?Void = g(x)?Void ∈ Type) supposing (g(x)?Void and f(x)?Void and f || g)
Proof
Definitions occuring in Statement :
fpf-compatible: f || g,
fpf-cap: f(x)?z,
fpf: a:A fp-> B[a],
deq: EqDecider(T),
uimplies: b supposing a,
uall: ∀[x:A]. B[x],
void: Void,
universe: Type,
equal: s = t ∈ T
Lemmas :
fpf-dom_wf,
subtype-fpf2,
top_wf,
subtype_top,
bool_wf,
fpf-ap_wf,
equal-wf-T-base,
assert_wf,
bnot_wf,
not_wf,
eqtt_to_assert,
uiff_transitivity,
eqff_to_assert,
assert_of_bnot
\mforall{}[X:Type]. \mforall{}[eq:EqDecider(X)]. \mforall{}[f,g:x:X fp-> Type]. \mforall{}[x:X].
(f(x)?Void = g(x)?Void) supposing (g(x)?Void and f(x)?Void and f || g)
Date html generated:
2015_07_17-AM-09_18_52
Last ObjectModification:
2015_01_28-AM-07_49_59
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