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Proof step 8 from proof natset-ls-max-ls (previous step)

n0 ∈ ns, max(m + 1, n) ≤ n0, ∀ n0. n0 ∈ ns → n0 < n ⊦ # ns = 0, ns = ∅

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: n0

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m < n → n ≠ 0; (s)% ⊦ ¬ max(n, m) < m; (s) Used terms: n0