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Proof step 1 from proof max-nls (conclusion)

⊦ ¬ m < (maxenc(m) * maxenc(m) + maxenc(m)) / 2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: 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-one: ⊦ m * m + m ≠ 1; from specification nat-mult%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