Nuprl Lemma : bag-combine-is-single-if

[A,B:Type]. ∀[f:A ⟶ bag(B)]. ∀[bs:bag(A)]. ∀[x:B].
  ⋃x∈bs.f[x] {x} ∈ bag(B) supposing ↓∃y:A. ((bs {y} ∈ bag(A)) ∧ (f[y] {x} ∈ bag(B)))


Proof




Definitions occuring in Statement :  bag-combine: x∈bs.f[x] single-bag: {x} bag: bag(T) uimplies: supposing a uall: [x:A]. B[x] so_apply: x[s] exists: x:A. B[x] squash: T and: P ∧ Q function: x:A ⟶ B[x] universe: Type equal: t ∈ T
Definitions unfolded in proof :  squash: T exists: x:A. B[x] and: P ∧ Q uall: [x:A]. B[x] member: t ∈ T so_lambda: λ2x.t[x] so_apply: x[s] prop: all: x:A. B[x] uimplies: supposing a
Lemmas referenced :  bag-combine-single-left equal_wf bag_wf bag-combine_wf squash_wf exists_wf single-bag_wf
Rules used in proof :  sqequalHypSubstitution sqequalSubstitution sqequalTransitivity computationStep sqequalReflexivity imageElimination cut productElimination thin hypothesis sqequalRule introduction extract_by_obid isectElimination hypothesisEquality because_Cache lambdaEquality applyEquality functionExtensionality cumulativity hyp_replacement equalitySymmetry applyLambdaEquality equalityTransitivity productEquality dependent_functionElimination functionEquality universeEquality isect_memberFormation isect_memberEquality axiomEquality

Latex:
\mforall{}[A,B:Type].  \mforall{}[f:A  {}\mrightarrow{}  bag(B)].  \mforall{}[bs:bag(A)].  \mforall{}[x:B].
    \mcup{}x\mmember{}bs.f[x]  =  \{x\}  supposing  \mdownarrow{}\mexists{}y:A.  ((bs  =  \{y\})  \mwedge{}  (f[y]  =  \{x\}))



Date html generated: 2017_10_01-AM-08_47_40
Last ObjectModification: 2017_07_26-PM-04_32_05

Theory : bags


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