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Proof step 3 from proof not-prime (previous step)

2 ≤ n, ∀ n0. n0 < n → ¬ prime(n0) ∧ 2 ≤ n0 → (∃ m. m | n0 ∧ prime(m) ∧ m * m ≤ n0) ⊦ n = 0, prime(n), ∃ m. m | n ∧ prime(m) ∧ m * m ≤ n

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ prime(n) ↔ 1 < n ∧ (∀ m. m | n → m = 1 ∨ m = n)
Proof step 5:
2 ≤ n, ∀ n0. n0 < n → ¬ prime(n0) ∧ 2 ≤ n0 → (∃ m. m | n0 ∧ prime(m) ∧ m * m ≤ n0) ⊦ n = 0, 1 < n ∧ (∀ m. m | n → m = 1 ∨ m = n), ∃ m. m | n ∧ prime(m) ∧ m * m ≤ n

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