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Step * 1 of Lemma bag-filter-union


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]} )
BY
{ Subst ⌜bag-union(bag-map(λb.[x∈b|p[x]];bs)) ~ [x∈bag-union(bs)|p[x]]⌝ 0⋅ }

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

2
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]] = [x∈bag-union(bs)|p[x]] ∈ bag({x:T| ↑p[x]} )

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Latex:

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Latex:

1.  T  :  Type
2.  p  :  T  {}\mrightarrow{}  \mBbbB{}
3.  as  :  bag(T)  List
4.  bs  :  bag(T)  List
5.  permutation(bag(T);as;bs)
\mvdash{}  [x\mmember{}bag-union(as)|p[x]]  =  bag-union(bag-map(\mlambda{}b.[x\mmember{}b|p[x]];bs))


By

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Latex:
Subst  \mkleeneopen{}bag-union(bag-map(\mlambda{}b.[x\mmember{}b|p[x]];bs))  \msim{}  [x\mmember{}bag-union(bs)|p[x]]\mkleeneclose{}  0\mcdot{}

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