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Proof step 4 from proof bag-add2-eq-elim (previous step)

∀ a0. (a0 = b ⊃ 1;0) + count(a0, bag0) = count(a0, bag1) ⊦ bag0 ++ b = bag1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ bag0 = bag1 ↔ (∀ a. count(a, bag0) = count(a, bag1))
Proof step 6:
∀ a0. (a0 = b ⊃ 1;0) + count(a0, bag0) = count(a0, bag1) ⊦ ∀ a. count(a, bag0 ++ b) = count(a, bag1)

Rule argument

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

Additional information

Used terms: bag1, bag0 ++ b