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Proof step 8 from proof max-rev (previous step)

¬((m0 * (m0 + 1)) / 2 ≤ m ∧ m < ((m0 + 1) * m0 + (m0 + 1) * 2) / 2 ∧ (m0 * (m0 + 1)) / 2 ≤ m), ¬ ((m0 + 1) * m0 + (m0 + 1) * 2) / 2 ≤ m, true, ∀ m2, m3, n. m2 - m3 < m - m0 → true → maxenc(m3, m2) = n ∧ (m3 * (m3 + 1)) / 2 ≤ m2 → (n * (n + 1)) / 2 ≤ m2 ∧ m2 < ((n + 1) * (n + 2)) / 2 ⊦ maxenc(m0, m) = n ∧ (m0 * (m0 + 1)) / 2 ≤ m → (n * (n + 1)) / 2 ≤ m ∧ m < ((n + 1) * (n + 2)) / 2

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: nat-div inst: name: div-add seq: ⊦ m0 % n = 0 ∧ 0 ≠ n → (m + m0) / n = m / n + m0 / n subst: [n, m0, m] → [2, (m0 + 1) * 2, (m0 + 1) * m0] flags: rewrite

Additional information

Used simplifier rules: rsucc-02: ⊦ m * (n + 1) = m * 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%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%mod-mult: n ≠ 0 ⊦ (m + n0 * n) % n = m % n; from specification nat-div%two-rec: ⊦ (m + 2) / 2 = m / 2 + 1; from specification nat-div%two-mdiv2-02: ⊦ (2 * m) / 2 = m; from specification nat-div%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%le-ls-simp: ⊦ m ≤ n → (n < m ↔ false); from specification nat%succ: ⊦ (* m0) * m + m = (m0 + 1) * m; from specification nat-mult%le-div-to-mult: m0 ≠ 0 ⊦ n ≤ m / m0 ↔ n * m0 ≤ m; from specification nat-div Used terms: 2, (m0 + 1) * 2, (m0 + 1) * m0