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Proof step 9 from proof prime-smallest (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
m ≠ 0 ⊦ m | n ↔ n % m = 0
Proof step 11:
n % m = 0, prime(m), m * m ≤ n, ∀ m. 2 ≤ m ∧ m * m ≤ n → n % m ≠ 0 ⊦ m = 0, n = 1, n = 0, prime(n)

Rule argument

spec: <toplevel> name: divi-mod seq: m ≠ 0 ⊦ m | n ↔ n % m = 0 subst: [n, m] → [n, m] flags: <all paths>

Additional information

Used simplifier rules: zero: ⊦ ¬ prime(0);% ⊦ 0 ≤ m; (s) Used terms: n, m