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Proof step 2 from proof max-nls (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
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-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult% ⊦ m * m + m ≠ 1; (s)% ⊦ m / 2 = 0 ↔ m = 0 ∨ m = 1; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic