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Proof step 1 from proof ex-smallest-notin (conclusion)

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

User applied rule apply lemma

And obtained the following new goals

Proof step 2:
⊦ ∃ n. ¬ n ∈ ns
Proof step 3:
∃ n. ¬ n ∈ ns ⊦ ∃ n. ¬ n ∈ ns ∧ (∀ m. ¬ m ∈ ns → n ≤ m)

Rule argument

spec: <toplevel> name: ex-notin seq: ⊦ ∃ n. ¬ n ∈ ns subst: [ns] → [ns]

Additional information

Used terms: ns