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Proof step 10 from proof max-snd-nls (previous step)

(maxenc(m) * maxenc(m) + maxenc(m)) % 2 = 0 ∧ ((maxenc(m) + 2) * maxenc(m) + maxenc(m)) % 2 = 0, 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) ≤ (maxenc(m) + 2) * maxenc(m) + maxenc(m) ⊦ (maxenc(m) + 2) * maxenc(m) + maxenc(m) = 0, maxenc(m) = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
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) + 2) * maxenc(m) + maxenc(m) = 0, maxenc(m) = 0

Rule argument

Additional information

Used simplifier rules: lep: ⊦ m ≤ m + n; from specification nat%le: ⊦ m0 ≠ 0 → (m0 * m ≤ m0 * n ↔ m ≤ n); from specification nat-mult%lep-02: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; from specification nat%square-even-03: ⊦ ((n + 2) * n + n) % 2 = 0; from specification nat-div%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%square-even-02: ⊦ (m * m + m) % 2 = 0; from specification nat-div