Nuprl Lemma : es-interface-val-disjoint
∀[Info:Type]. ∀[es:EO+(Info)]. ∀[A:Type]. ∀[Xs:EClass(A) List].
∀[X:EClass(A)]. ∀[e:E]. first-eclass(Xs)(e) = X(e) ∈ A supposing ↑e ∈b X supposing (X ∈ Xs)
supposing (∀X∈Xs.(∀Y∈Xs.(X = Y ∈ EClass(A)) ∨ X ∩ Y = 0))
Proof
Definitions occuring in Statement :
es-interface-disjoint: X ∩ Y = 0,
first-eclass: first-eclass(Xs),
eclass-val: X(e),
in-eclass: e ∈b X,
eclass: EClass(A[eo; e]),
event-ordering+: EO+(Info),
es-E: E,
l_all: (∀x∈L.P[x]),
l_member: (x ∈ l),
list: T List,
assert: ↑b,
uimplies: b supposing a,
uall: ∀[x:A]. B[x],
or: P ∨ Q,
universe: Type,
equal: s = t ∈ T
Lemmas :
first-eclass-val,
equal_wf,
squash_wf,
true_wf,
eclass-val_wf,
es-E_wf,
event-ordering+_subtype,
select_wf,
eclass_wf,
event-ordering+_wf,
sq_stable__le,
select_member
Latex:
\mforall{}[Info:Type]. \mforall{}[es:EO+(Info)]. \mforall{}[A:Type]. \mforall{}[Xs:EClass(A) List].
\mforall{}[X:EClass(A)]. \mforall{}[e:E]. first-eclass(Xs)(e) = X(e) supposing \muparrow{}e \mmember{}\msubb{} X supposing (X \mmember{} Xs)
supposing (\mforall{}X\mmember{}Xs.(\mforall{}Y\mmember{}Xs.(X = Y) \mvee{} X \mcap{} Y = 0))
Date html generated:
2015_07_20-PM-03_32_01
Last ObjectModification:
2015_01_27-PM-10_21_38
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