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Proof step 6 from proof subset-one (previous step)

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: a

Additional information

Used simplifier rules: ⊦ count(a, ∅ ++ a) = 1; (s)%not-zero: 0 < n ⊦ (* n) ≤ m → m ≠ 0; (forward) from specification nat Used terms: a