Nuprl Lemma : bag-combine-com

[A,B,C:Type]. ∀[f:A ⟶ B ⟶ bag(C)]. ∀[ba:bag(A)]. ∀[bb:bag(B)].  (⋃a∈ba.⋃b∈bb.f[a;b] = ⋃b∈bb.⋃a∈ba.f[a;b] ∈ bag(C))


Proof




Definitions occuring in Statement :  bag-combine: x∈bs.f[x] bag: bag(T) uall: [x:A]. B[x] so_apply: x[s1;s2] function: x:A ⟶ B[x] universe: Type equal: t ∈ T
Definitions unfolded in proof :  uall: [x:A]. B[x] member: t ∈ T bag: bag(T) quotient: x,y:A//B[x; y] and: P ∧ Q all: x:A. B[x] implies:  Q prop: subtype_rel: A ⊆B uimplies: supposing a so_lambda: λ2x.t[x] so_apply: x[s1;s2] so_apply: x[s] empty-bag: {} top: Top bag-append: as bs append: as bs so_lambda: so_lambda(x,y,z.t[x; y; z]) so_apply: x[s1;s2;s3] cand: c∧ B true: True squash: T guard: {T} iff: ⇐⇒ Q rev_implies:  Q so_lambda: λ2y.t[x; y]
Lemmas referenced :  bag_wf list_wf permutation_wf equal_wf equal-wf-base bag-combine_wf list-subtype-bag list_induction bag-combine-empty-left bag-combine-empty-right empty-bag_wf list_ind_cons_lemma list_ind_nil_lemma cons_wf nil_wf bag-combine-unit-left bag-combine-append-right bag-append_wf iff_weakening_equal squash_wf true_wf bag-combine-append-left quotient-member-eq permutation-equiv
Rules used in proof :  sqequalSubstitution sqequalTransitivity computationStep sqequalReflexivity isect_memberFormation introduction cut sqequalHypSubstitution pointwiseFunctionalityForEquality extract_by_obid isectElimination thin cumulativity hypothesisEquality hypothesis sqequalRule pertypeElimination productElimination equalityTransitivity equalitySymmetry lambdaFormation because_Cache rename dependent_functionElimination independent_functionElimination productEquality isect_memberEquality axiomEquality functionEquality hyp_replacement applyLambdaEquality applyEquality independent_isectElimination lambdaEquality functionExtensionality voidElimination voidEquality independent_pairFormation equalityUniverse levelHypothesis natural_numberEquality imageElimination imageMemberEquality baseClosed universeEquality

Latex:
\mforall{}[A,B,C:Type].  \mforall{}[f:A  {}\mrightarrow{}  B  {}\mrightarrow{}  bag(C)].  \mforall{}[ba:bag(A)].  \mforall{}[bb:bag(B)].
    (\mcup{}a\mmember{}ba.\mcup{}b\mmember{}bb.f[a;b]  =  \mcup{}b\mmember{}bb.\mcup{}a\mmember{}ba.f[a;b])



Date html generated: 2017_10_01-AM-08_47_24
Last ObjectModification: 2017_07_26-PM-04_31_57

Theory : bags


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