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Proof step 23 from proof max-zero-01 (previous step)

6 ≤ m, 3 ≤ m, 1 ≤ m, 3 ≤ maxenc(3, m) ⊦ maxenc(3, m) = 1 ↔ m = 1 ∨ m = 2, m = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
3 ≤ maxenc(3, m), 1 ≤ m, 3 ≤ m, 6 ≤ m ⊦ maxenc(3, m) = 1 ↔ m = 1 ∨ m = 2, m = 0, maxenc(3, m) = 0

Rule argument

Additional information

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