Nuprl Lemma : bag-union-as-combine

[A:Type]. ∀[x:bag(bag(A))].  (bag-union(x) = ⋃b∈x.b ∈ bag(A))


Proof




Definitions occuring in Statement :  bag-combine: x∈bs.f[x] bag-union: bag-union(bbs) bag: bag(T) uall: [x:A]. B[x] universe: Type equal: t ∈ T
Definitions unfolded in proof :  uall: [x:A]. B[x] member: t ∈ T bag-combine: x∈bs.f[x] squash: T prop: uimplies: supposing a all: x:A. B[x] true: True subtype_rel: A ⊆B guard: {T} iff: ⇐⇒ Q and: P ∧ Q rev_implies:  Q implies:  Q
Lemmas referenced :  equal_wf squash_wf true_wf bag_wf bag-union_wf bag-map-trivial subtype_rel_self iff_weakening_equal
Rules used in proof :  sqequalSubstitution sqequalTransitivity computationStep sqequalReflexivity isect_memberFormation introduction cut sqequalRule applyEquality thin lambdaEquality sqequalHypSubstitution imageElimination extract_by_obid isectElimination hypothesisEquality equalityTransitivity hypothesis equalitySymmetry universeEquality because_Cache independent_isectElimination lambdaFormation natural_numberEquality imageMemberEquality baseClosed instantiate productElimination independent_functionElimination isect_memberEquality axiomEquality

Latex:
\mforall{}[A:Type].  \mforall{}[x:bag(bag(A))].    (bag-union(x)  =  \mcup{}b\mmember{}x.b)



Date html generated: 2018_05_21-PM-06_24_14
Last ObjectModification: 2018_05_19-PM-05_15_11

Theory : bags


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