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Proof step 6 from proof square-interval (previous step)

(m * (m + 1)) % 2 = 0 ∧ 2 ≠ 0, m * (m + 1) ≤ ((n * (n + 1)) / 2 + n0) * 2, (m * (m + 1)) / 2 ≤ (n * (n + 1)) / 2 + n0, (n * (n + 1)) / 2 + n0 < ((m + 1) * (m + 2)) / 2, n0 ≤ n ⊦ m * (m + 1) ≤ n * (n + 1) + 2 * n0 ∧ n * (n + 1) + 2 * n0 < (m + 1) * (m + 2), m = n

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ m0 * (m + n) = m0 * m + m0 * n
Proof step 8:
(m * (m + 1)) % 2 = 0 ∧ 2 ≠ 0, m * (m + 1) ≤ 2 * (n * (n + 1)) / 2 + 2 * n0, (m * (m + 1)) / 2 ≤ (n * (n + 1)) / 2 + n0, (n * (n + 1)) / 2 + n0 < ((m + 1) * (m + 2)) / 2, n0 ≤ n ⊦ m * (m + 1) ≤ n * (n + 1) + 2 * n0 ∧ n * (n + 1) + 2 * n0 < (m + 1) * (m + 2), m = n

Rule argument

spec: nat-mult inst: name: dist seq: ⊦ m0 * (m + n) = m0 * m + m0 * n subst: [m0, n, m] → [2, n0, (n * (n + 1)) / 2] flags: rewrite

Additional information

Used simplifier rules: mult-c: ⊦ m * n = n * m; from specification nat-mult Used terms: 2, n0, (n * (n + 1)) / 2