Nuprl Lemma : bag-split

[T:Type]. ∀[p:T ⟶ 𝔹]. ∀[as:bag(T)].  (as ([x∈as|p[x]] [x∈as|¬bp[x]]) ∈ bag(T))


Proof




Definitions occuring in Statement :  bag-filter: [x∈b|p[x]] bag-append: as bs bag: bag(T) bnot: ¬bb bool: 𝔹 uall: [x:A]. B[x] so_apply: x[s] function: x:A ⟶ B[x] universe: Type equal: t ∈ T
Definitions unfolded in proof :  uall: [x:A]. B[x] member: t ∈ T squash: T prop: so_lambda: λ2x.t[x] so_apply: x[s] true: True subtype_rel: A ⊆B uimplies: supposing a guard: {T} iff: ⇐⇒ Q and: P ∧ Q rev_implies:  Q implies:  Q
Lemmas referenced :  equal_wf squash_wf true_wf bag_wf bag-filter-split iff_weakening_equal bool_wf
Rules used in proof :  sqequalSubstitution sqequalTransitivity computationStep sqequalReflexivity isect_memberFormation introduction cut applyEquality thin lambdaEquality sqequalHypSubstitution imageElimination extract_by_obid isectElimination hypothesisEquality equalityTransitivity hypothesis equalitySymmetry universeEquality sqequalRule functionExtensionality cumulativity because_Cache natural_numberEquality imageMemberEquality baseClosed independent_isectElimination productElimination independent_functionElimination isect_memberEquality axiomEquality functionEquality

Latex:
\mforall{}[T:Type].  \mforall{}[p:T  {}\mrightarrow{}  \mBbbB{}].  \mforall{}[as:bag(T)].    (as  =  ([x\mmember{}as|p[x]]  +  [x\mmember{}as|\mneg{}\msubb{}p[x]]))



Date html generated: 2017_10_01-AM-08_45_29
Last ObjectModification: 2017_07_26-PM-04_30_46

Theory : bags


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