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Proof step 7 from proof natset-ls-max-ls (previous step)

∀ n0. n0 ∈ ns → n0 < n ⊦ ∀ n0. n0 ∈ ns → n0 < max(m + 1, n)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
n0 ∈ ns, max(m + 1, n) ≤ n0, ∀ n0. n0 ∈ ns → n0 < n ⊦ # ns = 0, 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]% ⊦ m < n ↔ ¬ n ≤ m; (ss)