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Proof step 11 from proof max-in-ls (previous step)

m < n, bag0.max < n, n ∈ bag0 ++ m, bag0.max ≤ m, ∀ n0. ¬ bag0.max < n0 ∨ ¬ n0 ∈ bag0 ⊦ bag0 = ∅

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: n

Additional information

Used simplifier rules: ⊦ n ∈ bag0 ++ n0 ↔ n ∈ bag0 ∨ n = n0; (ws)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic% ⊦ m < n → ¬ n < m; (s) Used terms: n