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Proof step 2 from proof modprime (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ prime(n) ↔ 1 < n ∧ (∀ m. m | n → m = 1 ∨ m = n)
Proof step 4:
⊦ 1 < n ∧ (∀ m. m | n → m = 1 ∨ m = n) ↔ n ≠ 0 ∧ n ≠ 1 ∧ (∀ m. 2 ≤ m ∧ m < n → n % m ≠ 0)

Rule argument

spec: nat-prime inst: name: Prime seq: ⊦ prime(n) ↔ 1 < n ∧ (∀ m. m | n → m = 1 ∨ m = n) subst: [n] → [n] flags: rewrite, current

Additional information

Used terms: n