Nuprl Lemma : bag-append-ac

[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 :  uall: [x:A]. B[x] member: t ∈ T top: Top squash: T 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_wf bag-append-assoc equal_wf squash_wf true_wf bag-append_wf bag-append-comm iff_weakening_equal
Rules used in proof :  sqequalSubstitution sqequalTransitivity computationStep sqequalReflexivity isect_memberFormation introduction cut hypothesis extract_by_obid sqequalHypSubstitution isectElimination thin cumulativity hypothesisEquality sqequalRule isect_memberEquality axiomEquality because_Cache universeEquality voidElimination voidEquality applyEquality lambdaEquality imageElimination equalityTransitivity equalitySymmetry natural_numberEquality imageMemberEquality baseClosed independent_isectElimination productElimination independent_functionElimination

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_06
Last ObjectModification: 2017_07_26-PM-04_30_33

Theory : bags


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