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Proof step 3 from proof max-zero-01 (previous step)

⊦ maxenc(0, m) = 1 ↔ m = 1 ∨ m = 2

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ ((m + 1) * (m + 2)) / 2 ≤ n → maxenc(m, n) = maxenc(m + 1, n)
Proof step 5:
¬ ((0 + 1) * (0 + 2)) / 2 ≤ m ⊦ maxenc(0, m) = 1 ↔ m = 1 ∨ m = 2
Proof step 8:
((0 + 1) * (0 + 2)) / 2 ≤ m ⊦ maxenc(0 + 1, m) = 1 ↔ m = 1 ∨ m = 2

Rule argument

spec: nat-pair inst: name: maxi-rec seq: ⊦ ((m + 1) * (m + 2)) / 2 ≤ n → maxenc(m, n) = maxenc(m + 1, n) subst: [n, m] → [m, 0] flags: rewrite, current

Additional information

Used terms: m, 0