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Proof step 18 from proof max-def (previous step)

m0 < n, m < ((n + 1) * (n + 2)) / 2, m < ((m0 + 1) * (m0 + 2)) / 2, (n * (n + 1)) / 2 ≤ m, (m0 * (m0 + 1)) / 2 ≤ m, ∀ m2, m3, n. m2 - m3 < m - m0 → m2 < ((n + 1) * (n + 2)) / 2 ∧ (n * (n + 1)) / 2 ≤ m2 ∧ (m3 * (m3 + 1)) / 2 ≤ m2 → maxenc(m3, m2) = n ⊦ m0 = n

User applied rule insert spec-lemma

And obtained the following new goals

Proof step 19:
(m * (m + 1)) / 2 ≤ (n * (n + 1)) / 2 + n0, (n * (n + 1)) / 2 + n0 < ((m + 1) * (m + 2)) / 2, n0 ≤ n ⊦ m = n
Proof step 20:
m < ((m0 + 1) * (m0 + 2)) / 2, m < ((n + 1) * (n + 2)) / 2, m0 < n, (m0 * (m0 + 1)) / 2 ≤ m, (n * (n + 1)) / 2 ≤ m, ∀ m2, m3, n. m2 - m3 < m - m0 → m2 < ((n + 1) * (n + 2)) / 2 ∧ (n * (n + 1)) / 2 ≤ m2 ∧ (m3 * (m3 + 1)) / 2 ≤ m2 → maxenc(m3, m2) = n ⊦ m0 = n, (m0 * (m0 + 1)) / 2 ≤ (n * (n + 1)) / 2 + 0 ∧ (n * (n + 1)) / 2 + 0 < ((m0 + 1) * (m0 + 2)) / 2 ∧ 0 ≤ n

Rule argument

spec: nat-div name: square-interval seq: (m * (m + 1)) / 2 ≤ (n * (n + 1)) / 2 + n0, (n * (n + 1)) / 2 + n0 < ((m + 1) * (m + 2)) / 2, n0 ≤ n ⊦ m = n subst: [m, n, n0] → [m0, n, 0]

Additional information

Used terms: m0, n, 0