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

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

User applied rule insert elim lemma

And closed the goal!

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] → [n, n0 * m, m1, n2]

Additional information

Used simplifier rules: ⊦ n + 0 = n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m < n → n ≠ 0; (s)%n ≠ 0 ⊦ (m + n0 * n) % n = m % n; (s)%n ≠ 0 ⊦ (m * n) % n = 0; (s)%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult% ⊦ m < n → m % n = m; (s)% ⊦ m ≠ 0 → (m * n = 0 ↔ n = 0); (s)%n ≠ 0 ⊦ m0 = m / n ∧ n0 = m % n ↔ m = m0 * n + n0 ∧ n0 < n; (elim) Used terms: n, n0 * m, m1, n2