Nuprl Lemma : mapfilter-bor-eq
∀T,U:Type. ∀f:T ⟶ U. ∀P,Q:T ⟶ 𝔹. ∀L:T List.
((mapfilter(f;P;L) @ mapfilter(f;Q;L))
= (mapfilter(f;λx.(P[x] ∨bQ[x]);L) @ mapfilter(f;λx.(P[x] ∧b Q[x]);L))
∈ bag(U))
Proof
Definitions occuring in Statement :
mapfilter: mapfilter(f;P;L),
append: as @ bs,
list: T List,
bor: p ∨bq,
band: p ∧b q,
bool: 𝔹,
so_apply: x[s],
all: ∀x:A. B[x],
lambda: λx.A[x],
function: x:A ⟶ B[x],
universe: Type,
equal: s = t ∈ T,
bag: bag(T)
Definitions unfolded in proof :
all: ∀x:A. B[x],
uall: ∀[x:A]. B[x],
member: t ∈ T,
so_lambda: λ2x.t[x],
subtype_rel: A ⊆r B,
so_apply: x[s],
prop: ℙ,
uimplies: b supposing a,
implies: P ⇒ Q,
bool: 𝔹,
unit: Unit,
it: ⋅,
btrue: tt,
band: p ∧b q,
ifthenelse: if b then t else f fi ,
uiff: uiff(P;Q),
and: P ∧ Q,
bfalse: ff,
mapfilter: mapfilter(f;P;L),
top: Top,
append: as @ bs,
so_lambda: so_lambda(x,y,z.t[x; y; z]),
so_apply: x[s1;s2;s3],
or: P ∨ Q,
not: ¬A,
false: False,
cons-bag: x.b,
true: True,
iff: P ⇐⇒ Q,
rev_implies: P ⇐ Q,
squash: ↓T,
guard: {T},
bag-append: as + bs
Latex:
\mforall{}T,U:Type. \mforall{}f:T {}\mrightarrow{} U. \mforall{}P,Q:T {}\mrightarrow{} \mBbbB{}. \mforall{}L:T List.
((mapfilter(f;P;L) @ mapfilter(f;Q;L))
= (mapfilter(f;\mlambda{}x.(P[x] \mvee{}\msubb{}Q[x]);L) @ mapfilter(f;\mlambda{}x.(P[x] \mwedge{}\msubb{} Q[x]);L)))
Date html generated:
2016_05_17-AM-08_56_17
Last ObjectModification:
2016_01_17-PM-08_33_45
Theory : messages
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