Nuprl Lemma : bag-combine-combine
∀[A,B,C:Type]. ∀[b:bag(A)]. ∀[f:A ⟶ bag(B)]. ∀[g:B ⟶ bag(C)]. (⋃x∈⋃y∈b.f[y].g[x] = ⋃x∈b.⋃y∈f[x].g[y] ∈ bag(C))
Proof
Definitions occuring in Statement :
bag-combine: ⋃x∈bs.f[x],
bag: bag(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 :
uall: ∀[x:A]. B[x],
member: t ∈ T,
so_lambda: λ2x.t[x],
top: Top,
so_apply: x[s],
subtype_rel: A ⊆r B,
uimplies: b supposing a
Lemmas referenced :
bag-combine-assoc,
subtype_rel_bag,
top_wf,
trivial-equal,
bag_wf,
bag-combine_wf
Rules used in proof :
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
isect_memberFormation,
introduction,
cut,
equalitySymmetry,
sqequalRule,
lemma_by_obid,
sqequalHypSubstitution,
isectElimination,
thin,
isect_memberEquality,
voidElimination,
voidEquality,
hypothesisEquality,
applyEquality,
hypothesis,
independent_isectElimination,
lambdaEquality,
because_Cache,
functionEquality,
axiomEquality
Latex:
\mforall{}[A,B,C:Type]. \mforall{}[b:bag(A)]. \mforall{}[f:A {}\mrightarrow{} bag(B)]. \mforall{}[g:B {}\mrightarrow{} bag(C)].
(\mcup{}x\mmember{}\mcup{}y\mmember{}b.f[y].g[x] = \mcup{}x\mmember{}b.\mcup{}y\mmember{}f[x].g[y])
Date html generated:
2016_05_15-PM-02_29_08
Last ObjectModification:
2015_12_27-AM-09_49_51
Theory : bags
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