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Step * of Lemma bag-filter-union

∀[T:Type]. ∀[p:T ⟶ 𝔹]. ∀[bbs:bag(bag(T))].
  ([x∈bag-union(bbs)|p[x]] = bag-union(bag-map(λb.[x∈b|p[x]];bbs)) ∈ bag({x:T| ↑p[x]} ))
BY
{ ((Auto THEN BagD (-1)) THENA Auto) }

1
1. T : Type
2. p : T ⟶ 𝔹
3. as : bag(T) List
4. bs : bag(T) List
5. permutation(bag(T);as;bs)
⊢ [x∈bag-union(as)|p[x]] = bag-union(bag-map(λb.[x∈b|p[x]];bs)) ∈ bag({x:T| ↑p[x]} )

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Latex:

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Latex:
\mforall{}[T:Type].  \mforall{}[p:T  {}\mrightarrow{}  \mBbbB{}].  \mforall{}[bbs:bag(bag(T))].
    ([x\mmember{}bag-union(bbs)|p[x]]  =  bag-union(bag-map(\mlambda{}b.[x\mmember{}b|p[x]];bbs)))


By

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Latex:
((Auto  THEN  BagD  (-1))  THENA  Auto)

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