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Proof step 10 from proof subset-one (previous step)

⊦ count(a, bag) = 0, ∀ a0. count(a0, ∅ ++ a) ≤ count(a0, bag)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
count(a0, bag) < count(a0, ∅ ++ a) ⊦ count(a, bag) = 0

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m ≤ n ↔ ¬ n < m; (ss)