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Proof step 3 from proof mod-nle-01 (previous step)

n0 < m0, m0 * m + n0 ≤ m0 * n, n ≤ m ⊦ n0 = 0

User applied rule apply lemma

And obtained the following new goals

Proof step 4:
⊦ m0 < m ∧ n ≤ n0 ∧ n2 < n0 ∧ n1 < n → ¬ m * n0 + n2 ≤ m0 * n + n1

Rule argument

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

Additional information

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