Nuprl Lemma : hdf-base-ap
∀[A,B:Type]. ∀[F:A ⟶ bag(B)]. ∀[a:A]. (hdf-base(m.F[m])(a) = <hdf-base(m.F[m]), F[a]> ∈ (hdataflow(A;B) × bag(B)))
Proof
Definitions occuring in Statement :
hdf-base: hdf-base(m.F[m]),
hdf-ap: X(a),
hdataflow: hdataflow(A;B),
uall: ∀[x:A]. B[x],
so_apply: x[s],
function: x:A ⟶ B[x],
pair: <a, b>,
product: x:A × B[x],
universe: Type,
equal: s = t ∈ T,
bag: bag(T)
Definitions unfolded in proof :
uall: ∀[x:A]. B[x],
member: t ∈ T,
hdf-base: hdf-base(m.F[m]),
mk-hdf: mk-hdf(s,m.G[s; m];st.H[st];s0),
ifthenelse: if b then t else f fi ,
bfalse: ff,
top: Top,
so_lambda: λ2x.t[x],
so_apply: x[s]
Latex:
\mforall{}[A,B:Type]. \mforall{}[F:A {}\mrightarrow{} bag(B)]. \mforall{}[a:A]. (hdf-base(m.F[m])(a) = <hdf-base(m.F[m]), F[a]>)
Date html generated:
2016_05_16-AM-10_39_02
Last ObjectModification:
2015_12_28-PM-07_44_21
Theory : halting!dataflow
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