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Proof step 7 from proof ex-minimal-new-fd (previous step)

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

User applied rule apply induction

And closed the goal!

Rule argument

subst: [n0, ofh] → [m, ofh]

Additional information

Used terms: m, ofh