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Proof step 7 from proof ex-smallest-notin (previous step)

∀ n0, ns. n0 < n → true → n0 ∈ ns ∨ (∃ n. ¬ n ∈ ns ∧ (∀ m. ¬ m ∈ ns → n ≤ m)) ⊦ n ∈ ns, ∃ n. ¬ n ∈ ns ∧ (∀ m. ¬ m ∈ ns → n ≤ m), ¬ n ∈ ns ∧ (∀ m. ¬ m ∈ ns → n ≤ m)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
m < n, ∀ n0, ns. n0 < n → true → n0 ∈ ns ∨ (∃ n. ¬ n ∈ ns ∧ (∀ m. ¬ m ∈ ns → n ≤ m)) ⊦ n ∈ ns, m ∈ ns, ∃ n. ¬ n ∈ ns ∧ (∀ m. ¬ m ∈ ns → n ≤ m)

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m ≤ n ↔ ¬ n < m; (ss)