Nuprl Lemma : bag-combine-is-map

[A,B:Type]. ∀[b:bag(A)]. ∀[f:A ⟶ B].  (⋃x∈b.{f[x]} bag-map(f;b) ∈ bag(B))


Proof




Definitions occuring in Statement :  bag-combine: x∈bs.f[x] bag-map: bag-map(f;bs) single-bag: {x} bag: bag(T) 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 top: Top so_lambda: λ2x.t[x] so_apply: x[s] 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-combine-single-right-as-map equal_wf squash_wf true_wf bag_wf bag-map_wf eta_conv iff_weakening_equal
Rules used in proof :  sqequalSubstitution sqequalTransitivity computationStep sqequalReflexivity isect_memberFormation introduction cut sqequalRule extract_by_obid sqequalHypSubstitution isectElimination thin isect_memberEquality voidElimination voidEquality hypothesis applyEquality lambdaEquality imageElimination hypothesisEquality equalityTransitivity equalitySymmetry universeEquality because_Cache functionEquality cumulativity natural_numberEquality imageMemberEquality baseClosed functionExtensionality independent_isectElimination productElimination independent_functionElimination axiomEquality

Latex:
\mforall{}[A,B:Type].  \mforall{}[b:bag(A)].  \mforall{}[f:A  {}\mrightarrow{}  B].    (\mcup{}x\mmember{}b.\{f[x]\}  =  bag-map(f;b))



Date html generated: 2017_10_01-AM-08_47_47
Last ObjectModification: 2017_07_26-PM-04_32_10

Theory : bags


Home Index