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Proof step 2 from proof modls-02 (previous step)

n < n0, n2 < n0, m1 < m, n1 < n ⊦ m1 * n + n1 < m * n0 + n2

User applied rule apply lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: modls-01 seq: ⊦ n ≤ n0 ∧ n2 < n0 ∧ m1 < m ∧ n1 < n → m1 * n + n1 < m * n0 + n2 subst: [n, n0, n2, m1, m, n1] → [n, n0, n2, m1, m, n1]

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%trans-01: ⊦ n < n0 ∧ n0 < n1 → n +1 < n1; (forward) from specification nat-basic% ⊦ m < n → n ≠ 0; (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m < n → ¬ n < m; (s)%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 < 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: n, n0, n2, m1, m, n1