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

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

User applied rule apply lemma

And obtained the following new goals

Rule argument

spec: nat 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]

Additional information

Used simplifier rules: ls-06: ⊦ m < m0 ∧ n < n0 → m * n < m0 * n0;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%ls: ⊦ m0 * m < m0 * n ↔ m0 ≠ 0 ∧ m < n;%le: ⊦ m0 ≠ 0 → (m0 * m ≤ m0 * n ↔ m ≤ n);% ⊦ m + n < m + n0 ↔ n < n0; (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 ↔ ¬ n < m; (ss)% ⊦ m < n → ¬ n < m; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m < n → n ≠ 0; (s)% ⊦ ¬ n < n; (s) Used terms: m * m + m * m0, n * n + m * m0, n * n + n * m0