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Proof step 7 from proof modprime (previous step)

n % m = 0, 2 ≤ m, m < n, 2 < n, ∀ m. m | n → m = 1 ∨ m = n ⊦ n = 1, n = 0, m = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 8:
n = m0 * m + n0 ∧ n0 < m, n0 = 0, 2 ≤ m, m < n, 2 < n, ∀ m. m | n → m = 1 ∨ m = n ⊦ n = 1, n = 0, m = 0

Rule argument

spec: nat-div inst: name: div-elim seq: n ≠ 0 ⊦ m0 = m / n ∧ n0 = m % n ↔ m = m0 * n + n0 ∧ n0 < n subst: [m, n, m0, n0] → [n, m, m0, n0] flags:

Additional information

Used simplifier rules: n ≠ 0 ⊦ m0 = m / n ∧ n0 = m % n ↔ m = m0 * n + n0 ∧ n0 < n; (elim) Used terms: n, m, m0, n0