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Proof step 6 from proof mod-succ-01 (previous step)

n0 + 1 < n, n0 < n ⊦ ((m0 * n + n0) + 1) % n = n0 + 1, n0 + 1 = n

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Rule argument

spec: <toplevel> 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) + 1, n, m, n1] flags: <rotated> <all paths>

Additional information

Used simplifier rules: mod-eq: ⊦ m < n → m % n = m;% ⊦ n0 < n ∧ n1 < n → (m * n + n0 = m0 * n + n1 ↔ m = m0 ∧ n0 = n1); (ws)%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-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m + 1 ≠ 0; (s)% ⊦ 0 * n = 0; (s)% ⊦ m + n = m + n0 ↔ n = n0; (s) Used terms: (m0 * n + n0) + 1, n, m, n1