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Proof step 8 from proof align-down-add-ls (previous step)

(n0 + n1) / m * m < n0 ⊦ (n0 + n1) % m = 0, m = 0, n0 % m + n1 < m ∧ n0 % m ≠ 0

User applied rule insert elim lemma

And obtained the following new goals

Proof step 9:
n0 = m0 * m + n2, (n0 + n1) / m * m < n0, n2 < m ⊦ (n0 + n1) % m = 0, m = 0, n2 + n1 < m ∧ n2 ≠ 0

Rule argument

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

Additional information

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