Nuprl Lemma : classfun-eclass3
∀[Info,B,C:Type]. ∀[X:EClass(B ⟶ C)]. ∀[Y:EClass(B)]. ∀[es:EO+(Info)]. ∀[e:E].
(eclass3(X;Y)(e) = (X(e) Y(e)) ∈ C) supposing (X is functional and Y is functional)
Proof
Definitions occuring in Statement :
eclass3: eclass3(X;Y),
classfun: X(e),
es-functional-class: X is functional,
eclass: EClass(A[eo; e]),
event-ordering+: EO+(Info),
es-E: E,
uimplies: b supposing a,
uall: ∀[x:A]. B[x],
apply: f a,
function: x:A ⟶ B[x],
universe: Type,
equal: s = t ∈ T
Definitions unfolded in proof :
uall: ∀[x:A]. B[x],
member: t ∈ T,
uimplies: b supposing a,
classfun-res: X@e,
prop: ℙ,
subtype_rel: A ⊆r B,
so_lambda: λ2x y.t[x; y],
so_apply: x[s1;s2],
es-functional-class-at: X is functional at e,
and: P ∧ Q,
sq_type: SQType(T),
all: ∀x:A. B[x],
implies: P ⇒ Q,
guard: {T},
assert: ↑b,
ifthenelse: if b then t else f fi ,
btrue: tt,
true: True,
iff: P ⇐⇒ Q,
rev_implies: P ⇐ Q,
es-functional-class: X is functional
Latex:
\mforall{}[Info,B,C:Type]. \mforall{}[X:EClass(B {}\mrightarrow{} C)]. \mforall{}[Y:EClass(B)]. \mforall{}[es:EO+(Info)]. \mforall{}[e:E].
(eclass3(X;Y)(e) = (X(e) Y(e))) supposing (X is functional and Y is functional)
Date html generated:
2016_05_17-AM-11_16_50
Last ObjectModification:
2015_12_29-PM-05_11_35
Theory : process-model
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