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Proof step 6 from proof square-ls-03 (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 7:
(m * (m + 1)) / 2 + m0 = (m * (m + 1) + 2 * m0) / 2 ⊦ ¬ ((m + 1) * (m + 2)) / 2 ≤ ((n + 1) * (n + 2)) / 2, n < m, m0 ≤ m ∧ m ≤ n → (m * (m + 1)) / 2 + m0 < ((n + 1) * (n + 2)) / 2

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;%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%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2 Used formulas: (m * (m + 1)) / 2 + m0 = (m * (m + 1) + 2 * m0) / 2