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Proof step 5 from proof ls (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

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] → [m0 * n + n1, n0, m, n2] flags: rewrite, current

Additional information

Used simplifier rules: mod-neq-01: ⊦ m0 < m ∧ n ≤ n0 ∧ n2 < n0 ∧ n1 < n → m0 * n + n1 ≠ m * n0 + n2; 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%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: m0 * n + n1, n0, m, n2