Nuprl Lemma : fpf-compatible-join2
∀[A:Type]. ∀[eq:EqDecider(A)]. ∀[B:A ─→ Type]. ∀[f,g,h:a:A fp-> B[a]]. (f ⊕ g || h) supposing (g || h and f || h)
Proof
Definitions occuring in Statement :
fpf-join: f ⊕ g,
fpf-compatible: f || g,
fpf: a:A fp-> B[a],
deq: EqDecider(T),
uimplies: b supposing a,
uall: ∀[x:A]. B[x],
so_apply: x[s],
function: x:A ─→ B[x],
universe: Type
Lemmas :
fpf-compatible-symmetry,
fpf-join_wf,
fpf-compatible-join,
assert_wf,
fpf-dom_wf,
top_wf,
subtype-fpf2,
subtype_top,
fpf-compatible_wf,
fpf_wf,
deq_wf
\mforall{}[A:Type]. \mforall{}[eq:EqDecider(A)]. \mforall{}[B:A {}\mrightarrow{} Type]. \mforall{}[f,g,h:a:A fp-> B[a]].
(f \moplus{} g || h) supposing (g || h and f || h)
Date html generated:
2015_07_17-AM-11_12_34
Last ObjectModification:
2015_01_28-AM-07_42_02
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