### Nuprl Lemma : bag-combine-eq-right

`∀[A,B:Type]. ∀[b:bag(A)]. ∀[f1,f2:A ⟶ bag(B)].`
`  ⋃x∈b.f1[x] = ⋃x∈b.f2[x] ∈ bag(B) supposing ∀x:{x:A| x ↓∈ b} . (f1[x] = f2[x] ∈ bag(B))`

Proof

Definitions occuring in Statement :  bag-member: `x ↓∈ bs` bag-combine: `⋃x∈bs.f[x]` bag: `bag(T)` uimplies: `b supposing a` uall: `∀[x:A]. B[x]` so_apply: `x[s]` all: `∀x:A. B[x]` set: `{x:A| B[x]} ` function: `x:A ⟶ B[x]` universe: `Type` equal: `s = t ∈ T`
Definitions unfolded in proof :  bag-combine: `⋃x∈bs.f[x]` uall: `∀[x:A]. B[x]` member: `t ∈ T` squash: `↓T` exists: `∃x:A. B[x]` prop: `ℙ` so_lambda: `λ2x.t[x]` so_apply: `x[s]` bag-map: `bag-map(f;bs)` true: `True` uimplies: `b supposing a` subtype_rel: `A ⊆r B` guard: `{T}` iff: `P `⇐⇒` Q` and: `P ∧ Q` rev_implies: `P `` Q` implies: `P `` Q` all: `∀x:A. B[x]` nat: `ℕ` int_seg: `{i..j-}` lelt: `i ≤ j < k` le: `A ≤ B`
Lemmas referenced :  bag_to_squash_list all_wf bag-member_wf equal_wf bag_wf bag-map_wf bag-union_wf squash_wf true_wf iff_weakening_equal map_equal select_wf bag-member-select lelt_wf length_wf list-subtype-bag less_than_wf nat_wf length_wf_nat map_wf subtype_rel_self subtype_rel_set list_wf
Rules used in proof :  sqequalSubstitution sqequalTransitivity computationStep sqequalReflexivity cut introduction extract_by_obid sqequalHypSubstitution isectElimination thin because_Cache hypothesisEquality imageElimination productElimination promote_hyp hypothesis equalitySymmetry hyp_replacement applyLambdaEquality setEquality cumulativity sqequalRule lambdaEquality applyEquality functionExtensionality setElimination rename natural_numberEquality equalityTransitivity functionEquality universeEquality isect_memberFormation isect_memberEquality axiomEquality imageMemberEquality baseClosed independent_isectElimination independent_functionElimination lambdaFormation dependent_functionElimination dependent_set_memberEquality independent_pairFormation

Latex:
\mforall{}[A,B:Type].  \mforall{}[b:bag(A)].  \mforall{}[f1,f2:A  {}\mrightarrow{}  bag(B)].
\mcup{}x\mmember{}b.f1[x]  =  \mcup{}x\mmember{}b.f2[x]  supposing  \mforall{}x:\{x:A|  x  \mdownarrow{}\mmember{}  b\}  .  (f1[x]  =  f2[x])

Date html generated: 2017_10_01-AM-08_56_09
Last ObjectModification: 2017_07_26-PM-04_38_11

Theory : bags

Home Index