Nuprl Lemma : bag-append-assoc-comm

[T:Type]. ∀[as,bs,cs:bag(T)].  ((as bs cs) (bs as cs) ∈ bag(T))


Proof




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

Latex:
\mforall{}[T:Type].  \mforall{}[as,bs,cs:bag(T)].    ((as  +  bs  +  cs)  =  (bs  +  as  +  cs))



Date html generated: 2017_10_01-AM-08_45_03
Last ObjectModification: 2017_07_26-PM-04_30_29

Theory : bags


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