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Proof step 8 from proof ex-smallest-notin (
previous step
)
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)
User applied rule apply induction
And closed the goal!
Rule argument
subst: [n0, ns] → [m, ns]
Additional information
Used terms: m, ns