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Proof step 17 from proof square-ls-03 (previous step)

(m * (m + 1) + 2 * m0) % 2 = 0, ((m + 1) * (m + 2)) % 2 = 0, ((m + 1) * (m + 2)) / 2 ≤ ((n + 1) * (n + 2)) / 2, m ≤ n, m0 ≤ m ⊦ m * (m + 1) + 2 * m0 < (m + 1) * (m + 2), n < m, (m * (m + 1) + 2 * m0) / 2 < ((n + 1) * (n + 2)) / 2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
((m * m + m) + 2 * m0) % 2 = 0, ((n + 2) * n + n) / 2 < (m * m + m) / 2 + m0, m0 ≤ n, (m + 2) * m + 2 ≤ m * m + 2 * m0, ((m + 2) * m + m) / 2 * 2 ≤ (n + 2) * n + n, m ≤ n, m0 ≤ m ⊦

Rule argument

Additional information

Used simplifier rules: mod-once: n ≠ 0 ⊦ (m + n) % n = m % n;%square-even-03: ⊦ ((n + 2) * n + n) % 2 = 0;%two-rec: ⊦ (m + 2) / 2 = m / 2 + 1;%le-div-to-mult: m0 ≠ 0 ⊦ n ≤ m / m0 ↔ n * m0 ≤ m;%mult: n ≠ 0 ⊦ (n0 + m * n) / n = n0 / n + m;% ⊦ m < n → n ≠ 0; (s)%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ m ≤ n → (n < m ↔ false); (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m * (n + 1) = m * n + m; (s)