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Proof step 12 from proof max-zero-01 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 13:
⊦ ((m + 1) * (m + 2)) / 2 ≤ n → maxenc(m, n) = maxenc(m + 1, n)
Proof step 14:
((1 + 1) * (1 + 2)) / 2 ≤ m, 1 ≤ m, 3 ≤ m ⊦ maxenc(1 + 1, m) = 1 ↔ m = 1 ∨ m = 2, m = 0

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, 1] flags: rewrite, current

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%le-ls-simp: ⊦ m ≤ n → (n < m ↔ false); from specification nat Used terms: m, 1