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Proof step 15 from proof square-ls-03 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: mod-add seq: ⊦ m0 % n = 0 ∧ 0 ≠ n → (m + m0) % n = m % n subst: [n, m0, m] → [2, 2 * m0, m * (m + 1)] flags: <all paths>

Additional information

Used simplifier rules: 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;%mod-mult-01: n ≠ 0 ⊦ (m * n) % n = 0;%mod-mult-01: n ≠ 0 ⊦ (m * n) % n = 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;%square-even-02: ⊦ (m * m + m) % 2 = 0;%mod-once: n ≠ 0 ⊦ (m + n) % n = m % n;%square-even-03: ⊦ ((n + 2) * n + n) % 2 = 0;%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) = m * n + m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m ≤ n → (n < m ↔ false); (s) Used terms: 2, 2 * m0, m * (m + 1)