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Proof step 11 from proof square-interval (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 12:
((n * (n + 1)) / 2 + n0) * 2 < ((m + 1) * (m + 2)) / 2 * 2, (n * (n + 1)) % 2 = 0 ∧ 2 ≠ 0, (m * (m + 1)) % 2 = 0 ∧ 2 ≠ 0, m * (m + 1) ≤ n * (n + 1) + 2 * n0, (m * (m + 1)) / 2 ≤ (n * (n + 1)) / 2 + n0, (n * (n + 1)) / 2 + n0 < ((m + 1) * (m + 2)) / 2, n0 ≤ n ⊦ n * (n + 1) + 2 * n0 < (m + 1) * (m + 2), m = n

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;%rsucc-02: ⊦ m * (n + 1) = m * n + m; from specification nat-mult%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%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%coef-21: ⊦ m * n < m1 * n + n ↔ ¬ ¬(m < m1 + 1 ∧ n ≠ 0); from specification nat-mult Used formulas: ((n * (n + 1)) / 2 + n0) * 2 < ((m + 1) * (m + 2)) / 2 * 2