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Proof step 5 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)

User applied rule exists right

And obtained the following new goals

Proof step 6:
∀ 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)

Rule argument

pos: right 2 terms: n abort: F discard: F computed: T

Additional information

Used terms: n