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Proof step 1 from proof bag-diff-empty (conclusion)

⊦ bag \ bag = ∅

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ bag = bag2 ↔ (∀ n. count(n, bag) = count(n, bag2))
Proof step 3:
⊦ ∀ n. count(n, bag \ bag) = count(n, ∅)

Rule argument

spec: bag-basic, inst: nat-bag name: extension seq: ⊦ bag = bag2 ↔ (∀ n. count(n, bag) = count(n, bag2)) subst: [bag2, bag] → [∅, bag \ bag] flags: <all paths>

Additional information

Used terms: ∅, bag \ bag