Nuprl Lemma : Q-R-glued-conditional
∀[Info:Type]
∀es:EO+(Info)
∀[Q1,Q2,R:E ⟶ E ⟶ ℙ]. ∀[A,B:Type].
∀Ia1,Ia2:EClass(A). ∀Ib1,Ib2:EClass(B). ∀f:E([Ia1?Ia2]) ⟶ B.
(Ia1:Q1 →─f⟶ Ib1:R ⇒ Ia2:Q2 →─f⟶ Ib2:R ⇒ [Ia1?Ia2]:Q1|{Ia1} ∨ Q2|{Ia2} →─f⟶ [Ib1?Ib2]:R) supposing
(Ib1 ⋂ Ib2 = 0 and
Ia1 ⋂ Ia2 = 0)
Proof
Definitions occuring in Statement :
Q-R-glued: Ia:Qa →─f⟶ Ib:Rb,
es-interface-disjoint: X ⋂ Y = 0,
es-E-interface: E(X),
es-interface-predicate: {I},
cond-class: [X?Y],
eclass: EClass(A[eo; e]),
event-ordering+: EO+(Info),
es-E: E,
rel-restriction: R|P,
rel_or: R1 ∨ R2,
uimplies: b supposing a,
uall: ∀[x:A]. B[x],
prop: ℙ,
all: ∀x:A. B[x],
implies: P ⇒ Q,
function: x:A ⟶ B[x],
universe: Type
Definitions unfolded in proof :
so_apply: x[s],
so_lambda: λ2x.t[x],
exists: ∃x:A. B[x],
Q-R-glued: Ia:Qa →─f⟶ Ib:Rb,
top: Top,
so_apply: x[s1;s2],
so_lambda: λ2x y.t[x; y],
subtype_rel: A ⊆r B,
prop: ℙ,
and: P ∧ Q,
false: False,
implies: P ⇒ Q,
not: ¬A,
es-interface-disjoint: X ⋂ Y = 0,
member: t ∈ T,
uimplies: b supposing a,
all: ∀x:A. B[x],
uall: ∀[x:A]. B[x]
Latex:
\mforall{}[Info:Type]
\mforall{}es:EO+(Info)
\mforall{}[Q1,Q2,R:E {}\mrightarrow{} E {}\mrightarrow{} \mBbbP{}]. \mforall{}[A,B:Type].
\mforall{}Ia1,Ia2:EClass(A). \mforall{}Ib1,Ib2:EClass(B). \mforall{}f:E([Ia1?Ia2]) {}\mrightarrow{} B.
(Ia1:Q1 \mrightarrow{}{}f{}\mrightarrow{} Ib1:R
{}\mRightarrow{} Ia2:Q2 \mrightarrow{}{}f{}\mrightarrow{} Ib2:R
{}\mRightarrow{} [Ia1?Ia2]:Q1|\{Ia1\} \mvee{} Q2|\{Ia2\} \mrightarrow{}{}f{}\mrightarrow{} [Ib1?Ib2]:R) supposing
(Ib1 \mcap{} Ib2 = 0 and
Ia1 \mcap{} Ia2 = 0)
Date html generated:
2016_05_17-AM-07_56_41
Last ObjectModification:
2015_12_28-PM-11_26_05
Theory : event-ordering
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