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Step * 1 1 1 1 of Lemma bag-bind-com


1. A : Type
2. B : Type
3. C : Type
4. f : A ⟶ B ⟶ bag(C)
5. ba : bag(A)
⊢ bag-union(bag-map(λa.bag-union(bag-map(λb.f[a;b];[]));ba))
= bag-union(bag-map(λb.bag-union(bag-map(λa.f[a;b];ba));[]))
∈ bag(C)
BY
{ (RepUR ``bag-union bag-map`` 0
   THEN Folds ``bag-map empty-bag bag-union`` 0
   THEN Fold `bag-bind` 0
   THEN Auto
   THEN RWO "bag-bind-empty-right" 0
   THEN Auto) }

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Latex:

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Latex:

1.  A  :  Type
2.  B  :  Type
3.  C  :  Type
4.  f  :  A  {}\mrightarrow{}  B  {}\mrightarrow{}  bag(C)
5.  ba  :  bag(A)
\mvdash{}  bag-union(bag-map(\mlambda{}a.bag-union(bag-map(\mlambda{}b.f[a;b];[]));ba))
=  bag-union(bag-map(\mlambda{}b.bag-union(bag-map(\mlambda{}a.f[a;b];ba));[]))


By

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Latex:
(RepUR  ``bag-union  bag-map``  0
  THEN  Folds  ``bag-map  empty-bag  bag-union``  0
  THEN  Fold  `bag-bind`  0
  THEN  Auto
  THEN  RWO  "bag-bind-empty-right"  0
  THEN  Auto)

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