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Proof step 11 from proof max-rev (previous step)

((m0 + 1) * (m0 + 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 ⊦ m - m0 + 1 < m - m0, maxenc(m0 + 1, m) = n ∧ (m0 * (m0 + 1)) / 2 ≤ m → (n * (n + 1)) / 2 ≤ m ∧ m < ((n + 1) * (n + 2)) / 2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
(m0 * m0 + m0) / 2 ≤ m, ((m0 + 2) * m0 + m0) / 2 < m, m ≤ (m - m0 + 1) + m0, (m0 * m0 + m0) / 2 ≤ (m - m0 + 1) + m0, ((m0 + 2) * m0 + m0) / 2 < (m - m0 + 1) + m0, ∀ 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 ⊦ m = 0, m = 1, (maxenc(m0 + 1, m) * maxenc(m0 + 1, m) + maxenc(m0 + 1, m)) / 2 ≤ m ∧ ¬ ((maxenc(m0 + 1, m) + 2) * maxenc(m0 + 1, m) + maxenc(m0 + 1, m)) / 2 < m

Rule argument

Additional information

Used simplifier rules: eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%eq-02: n ≤ m ⊦ (m - n) + n0 = n1 ↔ m + n0 = n + n1; from specification nat%add-num-03: n < n0 ⊦ (* n) ≠ m0 + (* n0); from specification nat%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%ls-03: n0 ≤ n ⊦ n - n0 ≤ m ↔ n ≤ m + n0; from specification nat%ls-01: ⊦ m0 * m < m ↔ m0 = 0 ∧ m ≠ 0; from specification nat-mult%two-loup-12: ⊦ (m + n) / 2 < n ↔ m < n; from specification nat-div%trans-1: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic1%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%two-rec: ⊦ (m + 2) / 2 = m / 2 + 1; from specification nat-div%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%rsucc-02: ⊦ m * (n + 1) = m * n + m; from specification nat-mult