Nuprl Lemma : dataflow-valueall-type
∀[A:𝕌']. ∀[B:Type]. valueall-type(dataflow(A;B)) supposing ↓A
Proof
Definitions occuring in Statement :
dataflow: dataflow(A;B),
valueall-type: valueall-type(T),
uimplies: b supposing a,
uall: ∀[x:A]. B[x],
squash: ↓T,
universe: Type
Definitions unfolded in proof :
uall: ∀[x:A]. B[x],
member: t ∈ T,
uimplies: b supposing a,
dataflow: dataflow(A;B),
so_lambda: λ2x.t[x],
so_apply: x[s],
all: ∀x:A. B[x],
implies: P ⇒ Q,
prop: ℙ,
valueall-type: valueall-type(T),
has-value: (a)↓,
ext-eq: A ≡ B,
and: P ∧ Q,
subtype_rel: A ⊆r B,
exists: ∃x:A. B[x],
nat: ℕ,
le: A ≤ B,
less_than': less_than'(a;b),
false: False,
not: ¬A,
fun_exp: f^n,
top: Top,
compose: f o g,
squash: ↓T
Latex:
\mforall{}[A:\mBbbU{}']. \mforall{}[B:Type]. valueall-type(dataflow(A;B)) supposing \mdownarrow{}A
Date html generated:
2016_05_17-AM-10_19_21
Last ObjectModification:
2016_01_18-AM-00_20_47
Theory : process-model
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