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Proof step 6 from proof ex-minimal-new-fd (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
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)

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)