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Proof step 5 from proof max-zero (previous step)

¬ ¬ ((0 + 1) * (0 + 2)) / 2 ≤ m ⊦ maxenc(0, m) = 0 ↔ m = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
maxenc(0, m) = 0, 1 ≤ m ⊦ m = 0

Rule argument

Additional information

Used simplifier rules: not-zero: 0 < n ⊦ (* n) ≤ m → m ≠ 0; (forward) from specification nat%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