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Proof step 11 from proof square-ls-05 (previous step)

n * n + m * m0 < n * n + n * m0, m * m + m * m0 < n * n + m * m0, m < n, n * n + n * m0 ≤ m * m + m * m0 ⊦ n = 0

User applied rule insert spec-lemma

And obtained the following new goals

Rule argument

spec: nat inst: name: fle seq: ⊦ m < n ∧ n ≤ n0 → m < n0 subst: [m, n, n0] → [m * m + m * m0, n * n + m * m0, n * n + n * m0] flags:

Additional information

Used simplifier rules: ls: ⊦ m0 * m < m0 * n ↔ m0 ≠ 0 ∧ m < n;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%ls-06: ⊦ m < m0 ∧ n < n0 → m * n < m0 * n0;%le: ⊦ m0 ≠ 0 → (m0 * m ≤ m0 * n ↔ m ≤ n);%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%lep-02: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; from specification nat%lels-03: ⊦ m < n → m ≤ n; from specification nat%le-ls-simp: ⊦ m ≤ n → (n < m ↔ false); from specification nat Used terms: m * m + m * m0, n * n + m * m0, n * n + n * m0