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Proof step 7 from proof square-ls-03 (previous step)

((m + 1) * (m + 2)) / 2 ≤ ((n + 1) * (n + 2)) / 2, m ≤ n, m0 ≤ m ⊦ n < m, (m * (m + 1)) / 2 + m0 < ((n + 1) * (n + 2)) / 2

User applied rule cut formula

And obtained the following new goals

Proof step 8:
(m * (m + 1)) / 2 + m0 = (m * (m + 1) + 2 * m0) / 2, m0 ≤ m, m ≤ n, ((m + 1) * (m + 2)) / 2 ≤ ((n + 1) * (n + 2)) / 2 ⊦ (m * (m + 1)) / 2 + m0 < ((n + 1) * (n + 2)) / 2, n < m

Rule argument

fma: (m * (m + 1)) / 2 + m0 = (m * (m + 1) + 2 * m0) / 2

Additional information

Used simplifier rules: mult: n ≠ 0 ⊦ (n0 + m * n) / n = n0 / n + m;% ⊦ 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 = m + n0 ↔ n = n0; (s) Used formulas: (m * (m + 1)) / 2 + m0 = (m * (m + 1) + 2 * m0) / 2