Nuprl Lemma : fpf-sub-functionality2
∀[A,A':Type].
∀[B:A ─→ Type]. ∀[C:A' ─→ Type]. ∀[eq:EqDecider(A)]. ∀[eq':EqDecider(A')]. ∀[f,g:a:A fp-> B[a]].
(f ⊆ g) supposing (f ⊆ g and (∀a:A. (B[a] ⊆r C[a])))
supposing strong-subtype(A;A')
Proof
Definitions occuring in Statement :
fpf-sub: f ⊆ g,
fpf: a:A fp-> B[a],
deq: EqDecider(T),
strong-subtype: strong-subtype(A;B),
uimplies: b supposing a,
subtype_rel: A ⊆r B,
uall: ∀[x:A]. B[x],
so_apply: x[s],
all: ∀x:A. B[x],
function: x:A ─→ B[x],
universe: Type
Lemmas :
fpf-dom_functionality2,
subtype-fpf2,
top_wf,
subtype_top,
strong-subtype-deq-subtype,
fpf-ap_wf,
equal_wf
\mforall{}[A,A':Type].
\mforall{}[B:A {}\mrightarrow{} Type]. \mforall{}[C:A' {}\mrightarrow{} Type]. \mforall{}[eq:EqDecider(A)]. \mforall{}[eq':EqDecider(A')]. \mforall{}[f,g:a:A fp-> B[a]].
(f \msubseteq{} g) supposing (f \msubseteq{} g and (\mforall{}a:A. (B[a] \msubseteq{}r C[a])))
supposing strong-subtype(A;A')
Date html generated:
2015_07_17-AM-09_17_30
Last ObjectModification:
2015_01_28-AM-07_57_22
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