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Proof step 2 from proof mod-neq-01 (previous step)

m0 * n + n1 = m * n0 + n2, m0 < m, n2 < n0, n1 < n, n ≤ n0 ⊦

User applied rule apply lemma

And obtained the following new goals

Proof step 3:
⊦ 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] → [m0, m, n, n0, n2, n1]

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat% ⊦ m < n → n ≠ 0; (s)% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ ¬ n < n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: m0, m, n, n0, n2, n1