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Proof step 11 from proof subset-antisym (previous step)

count(a, bag0) < count(a, bag1), ∀ a. count(a, bag1) ≤ count(a, bag0), ∀ a. count(a, bag0) ≤ count(a, bag1) ⊦ count(a, bag0) = count(a, bag1), count(a, bag1) = 0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: a

Additional information

Used simplifier rules: le-ls-simp: ⊦ m ≤ n → (n < m ↔ false); from specification nat Used terms: a