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Proof step 3 from proof ls-06 (previous step)

m < m0, n < n0, m0 * n0 ≤ m * n ⊦

User applied rule apply lemma

And obtained the following new goals

Rule argument

spec: nat-basic name: trans seq: ⊦ n < n0 ∧ n0 < n1 → n < n1 subst: [n0, n, n1] → [m0 * n, m * n, m0 * n0]

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;%zero: ⊦ 0 * n = 0;%zero-01: ⊦ m0 * m = 0 ↔ m0 = 0 ∨ m = 0;% ⊦ m < n → n ≠ 0; (s)% ⊦ m ≤ 0 ↔ m = 0; (s)% ⊦ 0 < n ↔ n ≠ 0; (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% ⊦ ¬ n < n; (s) Used terms: m0 * n, m * n, m0 * n0