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Proof step 14 from proof square-ls-03 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: nat-mult inst: name: dist seq: ⊦ m0 * (m + n) = m0 * m + m0 * n subst: [m0, n, m] → [m, 2, m] flags: rewrite

Additional information

Used simplifier rules: succ: ⊦ (* m0) * m + m = (m0 + 1) * 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%lep-02: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; from specification nat%coef-18: ⊦ m * n + n ≤ m1 * n ↔ ¬ ¬(m + 1 ≤ m1 ∨ n = 0); from specification nat-mult%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; from specification nat-basic1%ref-1: ⊦ ¬ n < n; from specification nat-basic1%mult-c: ⊦ m * n = n * m; from specification nat-mult Used terms: m, 2, m