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Proof step 6 from proof is-aligned-is-aligned-mod (previous step)

(m0 * n0 + n1) % n = 0, n0 % n = 0, n1 < n0 ⊦ n1 % n = 0, n = 0

User applied rule insert elim lemma

And obtained the following new goals

Proof step 7:
n0 % n = 0, (m0 * n0 + n1) % n = 0, n1 = m1 * n + n2, n1 < n0, n2 < n ⊦ n = 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] → [n1, n, m, n2]

Additional information

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