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Proof step 24 from proof max-zero-01 (previous step)

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

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 2

Additional information

Used simplifier rules: lels-02: ⊦ m ≠ n → (m ≤ n ↔ m < n); from specification nat%one: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); from specification nat-basic1%not-zero: 0 < n ⊦ (* n) ≤ m → m ≠ 0; (forward) from specification nat