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Proof step 12 from proof square-interval (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 13:
(m * (m + 1)) % 2 = 0, (n * (n + 1)) % 2 = 0, (n * (n + 1)) / 2 + n0 < ((m + 1) * (m + 2)) / 2, ((n * (n + 1)) / 2 + n0) * 2 < ((m + 1) * (m + 2)) / 2 * 2, m * (m + 1) ≤ n * (n + 1) + 2 * n0, (m * (m + 1)) / 2 ≤ (n * (n + 1)) / 2 + n0, n0 ≤ n ⊦ 2 = 0, m = n, n * (n + 1) + 2 * n0 < (m + 1) * (m + 2)

Rule argument

fma: ((n * (n + 1)) / 2 + n0) * 2 < ((m + 1) * (m + 2)) / 2 * 2

Additional information

Used simplifier rules: square-even-02: ⊦ (m * m + m) % 2 = 0;%two-rec: ⊦ (m + 2) / 2 = m / 2 + 1;% ⊦ m * (n + 1) = m * n + m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m * n < m1 * n + n ↔ m < m1 + 1 ∧ n ≠ 0; (s) Used formulas: ((n * (n + 1)) / 2 + n0) * 2 < ((m + 1) * (m + 2)) / 2 * 2