Nuprl Lemma : fpf-cap_functionality_wrt_sub
∀[A:Type]. ∀[d1,d2,d3,d4:EqDecider(A)]. ∀[B:A ─→ Type]. ∀[f,g:a:A fp-> B[a]]. ∀[x:A]. ∀[z:B[x]].
(f(x)?z = g(x)?z ∈ B[x]) supposing ((↑x ∈ dom(f)) and f ⊆ g)
Proof
Definitions occuring in Statement :
fpf-sub: f ⊆ g,
fpf-cap: f(x)?z,
fpf-dom: x ∈ dom(f),
fpf: a:A fp-> B[a],
deq: EqDecider(T),
assert: ↑b,
uimplies: b supposing a,
uall: ∀[x:A]. B[x],
so_apply: x[s],
function: x:A ─→ B[x],
universe: Type,
equal: s = t ∈ T
Lemmas :
assert_wf,
fpf-dom_wf,
subtype-fpf2,
top_wf,
subtype_top,
fpf-sub_wf,
fpf_wf,
deq_wf,
bool_wf,
equal-wf-T-base,
bnot_wf,
not_wf,
eqtt_to_assert,
uiff_transitivity,
eqff_to_assert,
assert_of_bnot,
fpf-dom_functionality2
\mforall{}[A:Type]. \mforall{}[d1,d2,d3,d4:EqDecider(A)]. \mforall{}[B:A {}\mrightarrow{} Type]. \mforall{}[f,g:a:A fp-> B[a]]. \mforall{}[x:A]. \mforall{}[z:B[x]].
(f(x)?z = g(x)?z) supposing ((\muparrow{}x \mmember{} dom(f)) and f \msubseteq{} g)
Date html generated:
2015_07_17-AM-09_18_17
Last ObjectModification:
2015_01_28-AM-07_50_51
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