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Proof step 3 from proof exin (previous step)

⊦ count(a, bag) = 0, bag = (bag -- a) ++ a, ∃ bag0. bag = bag0 ++ a

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ bag0 = bag1 ↔ (∀ a. count(a, bag0) = count(a, bag1))
Proof step 5:
⊦ count(a, bag) = 0, ∀ a0. count(a0, bag) = count(a0, (bag -- a) ++ a), ∃ bag0. bag = bag0 ++ a

Rule argument

spec: bag-basic inst: name: extension seq: ⊦ bag0 = bag1 ↔ (∀ a. count(a, bag0) = count(a, bag1)) subst: [bag1, bag0] → [(bag -- a) ++ a, bag] flags: rewrite

Additional information

Used terms: (bag -- a) ++ a, bag