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Proof step 13 from proof max-snd-nls (previous step)

((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) * 2 < (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 14:
(maxenc(m) + (maxenc(m) * maxenc(m) + maxenc(m)) / 2) * 2 < (maxenc(m) + 2) * maxenc(m) + maxenc(m), m ≤ ((maxenc(m) + 2) * maxenc(m) + maxenc(m)) / 2, (maxenc(m) * maxenc(m) + maxenc(m)) / 2 ≤ m, maxenc(m) + (maxenc(m) * maxenc(m) + maxenc(m)) / 2 < m ⊦ maxenc(m) = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%f: ⊦ m < n → n ≠ 0; from specification nat-basic1%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%square-even-03: ⊦ ((n + 2) * n + n) % 2 = 0; from specification nat-div