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Proof step 7 from proof max-snd-nls (previous step)

(maxenc(m) * (maxenc(m) + 1)) / 2 ≤ m ∧ m < ((maxenc(m) + 1) * (maxenc(m) + 2)) / 2, maxenc(m) + (maxenc(m) * maxenc(m) + maxenc(m)) / 2 < m ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
maxenc(m) + (maxenc(m) * maxenc(m) + maxenc(m)) / 2 < m, (maxenc(m) * maxenc(m) + maxenc(m)) / 2 ≤ m, m ≤ ((maxenc(m) + 2) * maxenc(m) + maxenc(m)) / 2, maxenc(m) + (maxenc(m) * maxenc(m) + maxenc(m)) / 2 < ((maxenc(m) + 2) * maxenc(m) + maxenc(m)) / 2, (maxenc(m) * maxenc(m) + maxenc(m)) / 2 ≤ ((maxenc(m) + 2) * maxenc(m) + maxenc(m)) / 2 ⊦ (maxenc(m) + 2) * maxenc(m) + maxenc(m) = 0, maxenc(m) = 0

Rule argument

Additional information

Used simplifier rules: add-num-03: n < n0 ⊦ (* n) ≠ m0 + (* n0); from specification nat%zero-01: ⊦ m0 * m = 0 ↔ m0 = 0 ∨ m = 0; from specification nat-mult%z-01: ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; from specification nat-basic2%one: ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1; from specification nat-basic2%two-zero-01-01: ⊦ m / 2 = 0 ↔ ¬ ¬(m = 0 ∨ m = 1); from specification nat-div%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%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