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Proof step 1 from proof bag-swap-del (conclusion)

⊦ bag -- a -- b = bag -- b -- a

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ bag0 = bag1 ↔ (∀ a. count(a, bag0) = count(a, bag1))
Proof step 3:
⊦ ∀ a0. count(a0, bag -- a -- b) = count(a0, bag -- b -- a)

Rule argument

spec: bag-basic name: extension seq: ⊦ bag0 = bag1 ↔ (∀ a. count(a, bag0) = count(a, bag1)) subst: [bag1, bag0] → [bag -- b -- a, bag -- a -- b] flags: <all paths>

Additional information

Used terms: bag -- b -- a, bag -- a -- b