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Proof step 6 from proof two-03 (previous step)

m * m0 = 2 ⊦ m = 1, m = 2, m = 0

User applied rule apply lemma

And obtained the following new goals

Proof step 7:
⊦ m0 < m ∧ n ≤ n0 ∧ n2 < n0 ∧ n1 < n → m0 * n + n1 < m * n0 + n2

Rule argument

spec: <toplevel> name: mod-ls seq: ⊦ m0 < m ∧ n ≤ n0 ∧ n2 < n0 ∧ n1 < n → m0 * n + n1 < m * n0 + n2 subst: [m0, m, n, n0, n2, n1] → [2, m, 1, m0, 0, 0]

Additional information

Used simplifier rules: mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%mult-base: ⊦ m * 0 = 0;% ⊦ 0 < n ↔ n ≠ 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ 0 + m = m; (s)% ⊦ m ≠ n → (m < n ↔ ¬ n < m); (ss)% ⊦ m ≠ 1 → (m < 2 ↔ m = 0); (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m < 1 ↔ m = 0; (s) Used terms: 2, m, 1, m0, 0, 0