Nuprl Lemma : bag-filter-union2
∀[T:Type]. ∀[p:T ⟶ 𝔹]. ∀[bbs:bag(bag(T))]. ([x∈bag-union(bbs)|p[x]] = bag-union(bag-map(λb.[x∈b|p[x]];bbs)) ∈ bag(T))
Proof
Definitions occuring in Statement :
bag-union: bag-union(bbs),
bag-filter: [x∈b|p[x]],
bag-map: bag-map(f;bs),
bag: bag(T),
bool: 𝔹,
uall: ∀[x:A]. B[x],
so_apply: x[s],
lambda: λx.A[x],
function: x:A ⟶ B[x],
universe: Type,
equal: s = t ∈ T
Definitions unfolded in proof :
uall: ∀[x:A]. B[x],
member: t ∈ T,
so_lambda: λ2x.t[x],
so_apply: x[s],
subtype_rel: A ⊆r B,
prop: ℙ,
uimplies: b supposing a
Lemmas referenced :
bag-filter-union,
subtype_rel_bag,
assert_wf,
bool_wf
Rules used in proof :
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
isect_memberFormation,
introduction,
cut,
lemma_by_obid,
sqequalHypSubstitution,
isectElimination,
thin,
because_Cache,
sqequalRule,
hypothesisEquality,
hypothesis,
applyEquality,
setEquality,
independent_isectElimination,
lambdaEquality,
setElimination,
rename,
isect_memberEquality,
axiomEquality,
functionEquality
Latex:
\mforall{}[T:Type]. \mforall{}[p:T {}\mrightarrow{} \mBbbB{}]. \mforall{}[bbs:bag(bag(T))].
([x\mmember{}bag-union(bbs)|p[x]] = bag-union(bag-map(\mlambda{}b.[x\mmember{}b|p[x]];bbs)))
Date html generated:
2016_05_15-PM-02_27_10
Last ObjectModification:
2015_12_27-AM-09_51_33
Theory : bags
Home
Index