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Proof step 10 from proof align-down-add-ls (previous step)

n2 < m, ((n2 + n1) / m + m0) * m < m0 * m + n2, m ≤ n2 + n1 ⊦ (n2 + n1) % m = 0, n1 = 0, n2 = 0

User applied rule insert elim lemma

And obtained the following new goals

Proof step 11:
n2 + n1 = m1 * m + n0, (m1 + m0) * m < m0 * m + n2, n2 < m, n0 < m, m ≤ n2 + n1 ⊦ m = 0, n0 = 0, n1 = 0, 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] → [n2 + n1, m, m1, n]

Additional information

Used simplifier rules: ⊦ 0 ≤ m; (s)% ⊦ ¬ m < 0; (s)%n ≠ 0 ⊦ m0 = m / n ∧ n0 = m % n ↔ m = m0 * n + n0 ∧ n0 < n; (elim) Used terms: n2 + n1, m, m1, n