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Proof step 2 from proof some (previous step)

⊦ bag = ∅, ∃ a. count(a, bag) ≠ 0

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:
⊦ ∀ a. count(a, bag) = count(a, ∅), ∃ a. count(a, bag) ≠ 0

Rule argument

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

Additional information

Used terms: ∅, bag