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Proof step 5 from proof modprime (previous step)

⊦ (∀ m. m | n → m = 1 ∨ m = n) ↔ (∀ m. 2 ≤ m ∧ m < n → n % m ≠ 0), n = 1, n = 0

User applied rule case distinction

And obtained the following new goals

Proof step 6:
∀ m. m | n → m = 1 ∨ m = n ⊦ ∀ m. 2 ≤ m ∧ m < n → n % m ≠ 0, n = 1, n = 0
Proof step 10:
∀ m. 2 ≤ m ∧ m < n → n % m ≠ 0 ⊦ ∀ m. m | n → m = 1 ∨ m = n, n = 1, n = 0

Rule argument

pos: right 1

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