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Proof step 2 from proof mod-succ-01 (previous step)

⊦ (m + 1) % n = m % n + 1, m % n + 1 = n, n = 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] → [m, n, m0, n0] flags: rewrite, current

Additional information

Used simplifier rules: mod-eq-01: ⊦ m0 < n ∧ n ≠ 0 → ((m + m0) % n = m0 ↔ m % n = 0);%mod-mult-01: n ≠ 0 ⊦ (m * n) % n = 0;%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%succ-03: ⊦ m < n ∧ m + 1 ≠ n → m + 1 < n; (forward) from specification nat-basic2%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult Used terms: m, n, m0, n0