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Proof step 13 from proof natset-ls-add-max-ls (previous step)

m + 1 ≤ n0, ∀ n0. n0 ∈ ns → n0 < n, n0 = m ∨ n0 ∈ ns ⊦ m + 1 < n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
m < n0, n0 ∈ ns, n ≤ m + 1, n ≤ n0, ∀ n0. n0 ∈ ns → n0 < n ⊦ # ns = 0, n0 = m, ns = ∅

Rule argument

Additional information

Used simplifier rules: size-zero-01: ⊦ ns ≠ ∅ → # ns ≠ 0; (forward) from specification set-basic[nat-set]%ne-01: ⊦ n ∈ ns → ns ≠ ∅; (forward) from specification set-basic[nat-set]% ⊦ ¬ n < n; (s)% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%le-one: ⊦ m ≤ n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat% ⊦ m < n ↔ ¬ n ≤ m; (ss)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)