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Proof step 4 from proof square-interval (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ n % m = 0 ∧ m ≠ 0 → m * n / m = n
Proof step 6:
(m * (m + 1)) % 2 = 0 ∧ 2 ≠ 0, m * (m + 1) ≤ ((n * (n + 1)) / 2 + n0) * 2, (m * (m + 1)) / 2 ≤ (n * (n + 1)) / 2 + n0, (n * (n + 1)) / 2 + n0 < ((m + 1) * (m + 2)) / 2, n0 ≤ n ⊦ m * (m + 1) ≤ n * (n + 1) + 2 * n0 ∧ n * (n + 1) + 2 * n0 < (m + 1) * (m + 2), m = n

Rule argument

spec: nat-div inst: name: mult-div seq: ⊦ n % m = 0 ∧ m ≠ 0 → m * n / m = n subst: [m, n] → [2, m * (m + 1)] flags: rewrite, current

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;%square-even-02: ⊦ (m * m + m) % 2 = 0;%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%le-num: 0 ≠ m ⊦ (* m) * n ≤ m * n0 ↔ n ≤ n0; from specification nat-mult%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%mult-c: ⊦ m * n = n * m; from specification nat-mult Used terms: 2, m * (m + 1)