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Proof step 6 from proof max-def (previous step)

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

User applied rule insert spec-lemma

And obtained the following new goals

Proof step 7:
(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 8:
¬ ((m0 + 1) * (m0 + 2)) / 2 ≤ m, ¬ ((m0 + 1) * (m0 + 2)) / 2 ≤ m, true, ∀ m2, m3, n. m2 - m3 < m - m0 → true → (n * (n + 1)) / 2 ≤ m2 ∧ m2 < ((n + 1) * (n + 2)) / 2 ∧ (m3 * (m3 + 1)) / 2 ≤ m2 → maxenc(m3, m2) = n ⊦ (n * (n + 1)) / 2 ≤ (m0 * (m0 + 1)) / 2 + 0 ∧ (m0 * (m0 + 1)) / 2 + 0 < ((n + 1) * (n + 2)) / 2 ∧ 0 ≤ m0, (n * (n + 1)) / 2 ≤ m ∧ m < ((n + 1) * (n + 2)) / 2 ∧ (m0 * (m0 + 1)) / 2 ≤ m → m0 = n

Rule argument

spec: nat-div inst: 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] → [n, m0, 0] flags:

Additional information

Used terms: n, m0, 0