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Proof step 2 from proof diff-add-sw (previous step)

⊦ (bag0 ++ a) \ bag1 = (bag0 \ bag1) ++ a, a ∈ bag1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ bag0 = bag1 ↔ (∀ a. count(a, bag0) = count(a, bag1))
Proof step 4:
⊦ ∀ a0. count(a0, (bag0 ++ a) \ bag1) = count(a0, (bag0 \ bag1) ++ a), a ∈ bag1

Rule argument

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

Additional information

Used terms: (bag0 \ bag1) ++ a, (bag0 ++ a) \ bag1