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Proof step 14 from proof cong-equiv (previous step)

n1 + n0 = m0 * m + n0 % m, n0 % m < m ⊦ n1 = 0, m | n1

User applied rule insert elim lemma

And obtained the following new goals

Proof step 15:
n0 = m1 * m + n2 ∧ n2 < m ∧ m ≠ 0, n1 + n0 = m0 * m + n2, n2 < m ⊦ n1 = 0, m | n1

Rule argument

spec: nat-div inst: 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, m1, n] flags:

Additional information

Used simplifier rules: mult-base: ⊦ m * 0 = 0; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%z: ⊦ 0 + m = m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%less-1: ⊦ ¬ n < 0; from specification nat-basic1%n ≠ 0 ⊦ m0 = m / n ∧ n0 = m % n ↔ m = m0 * n + n0 ∧ n0 < n; (elim) Used terms: n0, m, m1, n