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Proof step 8 from proof max-zero-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
1 ≤ m ⊦ maxenc(1, m) = 1 ↔ m = 1 ∨ m = 2, 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