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Proof step 4 from proof is-aligned-of-mod (previous step)

m0 % m1 = 0 ⊦ (n % m0) % m1 = 0 ↔ n % m1 = 0, m0 = 0, m1 = 0

User applied rule insert elim lemma

And obtained the following new goals

Proof step 5:
m0 = m2 * m1 + n0, n0 = 0, n0 < m1 ⊦ (n % m0) % m1 = 0 ↔ n % m1 = 0, m0 = 0, m1 = 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] → [m0, m1, m, n0]

Additional information

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