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Proof step 3 from proof ex-smallest-notin (previous step)

∃ n. ¬ n ∈ ns ⊦ ∃ n. ¬ n ∈ ns ∧ (∀ m. ¬ m ∈ ns → n ≤ m)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ n ∈ ns, ∃ n. ¬ n ∈ ns ∧ (∀ m. ¬ m ∈ ns → n ≤ m)

Rule argument

Additional information