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Proof step 5 from proof le-succ (previous step)

m0 < (m0 * n + n0) / (n + 1), n0 < n ⊦

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 + n0, n + 1, m, n1] flags: rewrite, current

Additional information

Used simplifier rules: dist: ⊦ m0 * (m + n) = m0 * m + m0 * n; from specification nat-mult%one: ⊦ m0 * 1 = m0; 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%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%mod-neq: ⊦ m0 < m ∧ n0 < n ∧ n1 ≤ n → m0 * n + n0 ≠ m * n + m + n1; from specification nat-mult Used terms: m0 * n + n0, n + 1, m, n1