Nuprl Lemma : fpf-sub-join-right2
∀[A:Type]. ∀[B,C:A ⟶ Type]. ∀[eq:EqDecider(A)]. ∀[f:a:A fp-> B[a]]. ∀[g:a:A fp-> C[a]].
g ⊆ f ⊕ g supposing ∀x:A. (((↑x ∈ dom(f)) ∧ (↑x ∈ dom(g))) ⇒ ((B[x] ⊆r C[x]) c∧ (f(x) = g(x) ∈ C[x])))
Proof
Definitions occuring in Statement :
fpf-join: f ⊕ g,
fpf-sub: f ⊆ g,
fpf-ap: f(x),
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],
cand: A c∧ B,
so_apply: x[s],
all: ∀x:A. B[x],
implies: P ⇒ Q,
and: P ∧ Q,
function: x:A ⟶ B[x],
universe: Type,
equal: s = t ∈ T
Definitions unfolded in proof :
fpf-sub: f ⊆ g,
uall: ∀[x:A]. B[x],
uimplies: b supposing a,
member: t ∈ T,
all: ∀x:A. B[x],
implies: P ⇒ Q,
cand: A c∧ B,
so_lambda: λ2x.t[x],
so_apply: x[s],
subtype_rel: A ⊆r B,
top: Top,
prop: ℙ,
and: P ∧ Q,
iff: P ⇐⇒ Q,
rev_implies: P ⇐ Q,
guard: {T},
or: P ∨ Q,
fpf-join: f ⊕ g,
fpf-ap: f(x),
pi2: snd(t),
fpf-cap: f(x)?z,
bool: 𝔹,
unit: Unit,
it: ⋅,
btrue: tt,
uiff: uiff(P;Q),
ifthenelse: if b then t else f fi ,
bfalse: ff
Latex:
\mforall{}[A:Type]. \mforall{}[B,C:A {}\mrightarrow{} Type]. \mforall{}[eq:EqDecider(A)]. \mforall{}[f:a:A fp-> B[a]]. \mforall{}[g:a:A fp-> C[a]].
g \msubseteq{} f \moplus{} g supposing \mforall{}x:A. (((\muparrow{}x \mmember{} dom(f)) \mwedge{} (\muparrow{}x \mmember{} dom(g))) {}\mRightarrow{} ((B[x] \msubseteq{}r C[x]) c\mwedge{} (f(x) = g(x))))
Date html generated:
2016_05_16-AM-11_11_57
Last ObjectModification:
2015_12_29-AM-09_19_51
Theory : event-ordering
Home
Index