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Proof step 4 from proof modprime (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⊦ (∀ m. m | n → m = 1 ∨ m = n) ↔ (∀ m. 2 ≤ m ∧ m < n → n % m ≠ 0), n = 1, n = 0

Rule argument

Additional information

Used simplifier rules: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); (s)