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Proof step 2 from proof mod-neq (previous step)

m0 * n + n0 = m * n + m + n1, m0 < m, n0 < n, n1 ≤ n ⊦

User applied rule apply lemma

And obtained the following new goals

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] → [m0, m, n, n + 1, n1, n0]

Additional information

Used simplifier rules: dist: ⊦ m0 * (m + n) = m0 * m + m0 * n;%one: ⊦ m0 * 1 = m0;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;% ⊦ m < n + 1 ↔ ¬ n < m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n → n ≠ 0; (s)% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ ¬ n < n; (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ ¬ m + n < m; (s)%trans-01: ⊦ n < n0 ∧ n0 < n1 → n +1 < n1; (forward) from specification nat-basic%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat Used terms: m0, m, n, n + 1, n1, n0