Nuprl Lemma : fpf-join-wf
∀[A:Type]. ∀[B,C,D:A ⟶ Type]. ∀[f:a:A fp-> B[a]]. ∀[g:a:A fp-> C[a]]. ∀[eq:EqDecider(A)].
(f ⊕ g ∈ a:A fp-> D[a]) supposing
((∀a:A. ((↑a ∈ dom(g)) ⇒ (C[a] ⊆r D[a]))) and
(∀a:A. ((↑a ∈ dom(f)) ⇒ (B[a] ⊆r D[a]))))
Proof
Definitions occuring in Statement :
fpf-join: f ⊕ g,
fpf-dom: x ∈ dom(f),
fpf: a:A fp-> B[a],
deq: EqDecider(T),
assert: ↑b,
uimplies: b supposing a,
subtype_rel: A ⊆r B,
uall: ∀[x:A]. B[x],
so_apply: x[s],
all: ∀x:A. B[x],
implies: P ⇒ Q,
member: t ∈ T,
function: x:A ⟶ B[x],
universe: Type
Definitions unfolded in proof :
fpf: a:A fp-> B[a],
fpf-join: f ⊕ g,
pi1: fst(t),
all: ∀x:A. B[x],
member: t ∈ T,
top: Top,
fpf-cap: f(x)?z,
fpf-dom: x ∈ dom(f),
uall: ∀[x:A]. B[x],
prop: ℙ,
implies: P ⇒ Q,
bool: 𝔹,
unit: Unit,
it: ⋅,
btrue: tt,
ifthenelse: if b then t else f fi ,
uiff: uiff(P;Q),
and: P ∧ Q,
uimplies: b supposing a,
iff: P ⇐⇒ Q,
subtype_rel: A ⊆r B,
so_apply: x[s],
rev_implies: P ⇐ Q,
bfalse: ff,
exists: ∃x:A. B[x],
or: P ∨ Q,
sq_type: SQType(T),
guard: {T},
bnot: ¬bb,
assert: ↑b,
false: False,
not: ¬A,
so_lambda: λ2x.t[x]
Latex:
\mforall{}[A:Type]. \mforall{}[B,C,D:A {}\mrightarrow{} Type]. \mforall{}[f:a:A fp-> B[a]]. \mforall{}[g:a:A fp-> C[a]]. \mforall{}[eq:EqDecider(A)].
(f \moplus{} g \mmember{} a:A fp-> D[a]) supposing
((\mforall{}a:A. ((\muparrow{}a \mmember{} dom(g)) {}\mRightarrow{} (C[a] \msubseteq{}r D[a]))) and
(\mforall{}a:A. ((\muparrow{}a \mmember{} dom(f)) {}\mRightarrow{} (B[a] \msubseteq{}r D[a]))))
Date html generated:
2016_05_16-AM-11_09_34
Last ObjectModification:
2015_12_29-AM-09_28_29
Theory : event-ordering
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