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Proof step 2 from proof prime-smallest (previous step)

⊦ prime(n) ↔ n ≠ 0 ∧ n ≠ 1 ∧ (∀ m. 2 ≤ m ∧ m * m ≤ n → n % m ≠ 0)

User applied rule case distinction

And obtained the following new goals

Proof step 3:
prime(n) ⊦ n ≠ 0 ∧ n ≠ 1 ∧ (∀ m. 2 ≤ m ∧ m * m ≤ n → n % m ≠ 0)
Proof step 5:
∀ m. 2 ≤ m ∧ m * m ≤ n → n % m ≠ 0 ⊦ n = 1, n = 0, prime(n)

Rule argument

pos: right 1

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