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Proof step 16 from proof max-snd-nls (previous step)

2 * maxenc(m) + 2 * (maxenc(m) * maxenc(m) + maxenc(m)) / 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
maxenc(m) + 2 * (maxenc(m) * maxenc(m) + maxenc(m)) / 2 < (maxenc(m) + 2) * maxenc(m), 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) = 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%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%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%one: ⊦ m0 * 1 = m0; from specification nat-mult%numeral-21: m2 ≠ 0 ⊦ (* m2) * m + n < m + n0 ↔ (m2 - 1) * m + n < n0; from specification nat-mult