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
Definitions unfolded in proof :
uall: ∀[x:A]. B[x],
member: t ∈ T,
uimplies: b supposing a,
so_lambda: λ2x.t[x],
so_apply: x[s],
fpf-compatible: f || g,
all: ∀x:A. B[x],
implies: P ⇒ Q,
and: P ∧ Q,
subtype_rel: A ⊆r B,
top: Top,
prop: ℙ
Latex:
\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:
2016_05_16-AM-11_29_08
Last ObjectModification:
2015_12_29-AM-09_25_11
Theory : event-ordering
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