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Proof step 8 from proof div-two-mult (previous step)

(m0 + 1) % 2 = 1, m0 % 2 = 0 ⊦ 2 * (m0 + 1) / 2 = m0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
n ≠ 0 ⊦ m % n + 1 ≠ n → (m + 1) / n = m / n

Rule argument

spec: <toplevel> name: div-succ-01 seq: n ≠ 0 ⊦ m % n + 1 ≠ n → (m + 1) / n = m / n subst: [n, m] → [2, m0] flags: <all paths>

Additional information

Used simplifier rules: two-odd-01: ⊦ m % 2 = 1 ↔ m % 2 ≠ 0;%mult-02: n ≠ 0 ⊦ m % n = 0 → m / n * n = m;%n0 < n ⊦ (* n) = m0 + (* n0) ↔ n - n0 = m0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult Used terms: 2, m0