Nuprl Lemma : fpf-join-assoc
∀[A:Type]. ∀[B:A ⟶ Type]. ∀[eq:EqDecider(A)]. ∀[f,g,h:a:A fp-> B[a]]. (f ⊕ g ⊕ h = f ⊕ g ⊕ h ∈ a:A fp-> B[a])
Proof
Definitions occuring in Statement :
fpf-join: f ⊕ g,
fpf: a:A fp-> B[a],
deq: EqDecider(T),
uall: ∀[x:A]. B[x],
so_apply: x[s],
function: x:A ⟶ B[x],
universe: Type,
equal: s = t ∈ T
Definitions unfolded in proof :
fpf-join: f ⊕ g,
fpf: a:A fp-> B[a],
fpf-ap: f(x),
fpf-cap: f(x)?z,
fpf-dom: x ∈ dom(f),
uall: ∀[x:A]. B[x],
member: t ∈ T,
pi1: fst(t),
pi2: snd(t),
prop: ℙ,
so_apply: x[s],
top: Top,
squash: ↓T,
all: ∀x:A. B[x],
true: True,
iff: P ⇐⇒ Q,
and: P ∧ Q,
not: ¬A,
implies: P ⇒ Q,
rev_implies: P ⇐ Q,
or: P ∨ Q,
false: False,
bool: 𝔹,
unit: Unit,
it: ⋅,
btrue: tt,
uiff: uiff(P;Q),
uimplies: b supposing a,
bfalse: ff,
exists: ∃x:A. B[x],
sq_type: SQType(T),
guard: {T},
bnot: ¬bb,
ifthenelse: if b then t else f fi ,
assert: ↑b,
cand: A c∧ B,
so_lambda: λ2x.t[x]
Latex:
\mforall{}[A:Type]. \mforall{}[B:A {}\mrightarrow{} Type]. \mforall{}[eq:EqDecider(A)]. \mforall{}[f,g,h:a:A fp-> B[a]]. (f \moplus{} g \moplus{} h = f \moplus{} g \moplus{} h)
Date html generated:
2016_05_16-AM-11_10_25
Last ObjectModification:
2016_01_17-PM-03_52_35
Theory : event-ordering
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