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Proof step 5 from proof subset-in (previous step)

count(a, bag1) = 0, ∀ a0. count(a0, bag0 ++ a) ≤ count(a0, bag1) ⊦

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, bag ++ a) = count(a, bag) + 1; (s)% ⊦ m + 1 ≤ n ↔ m < n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m < n → n ≠ 0; (s) Used terms: a