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

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

User applied rule insert lemma

And obtained the following new goals

Proof step 3:
⊦ m0 < m ∧ n ≤ n0 ∧ n2 < n0 ∧ n1 < n → ¬ m * n0 + n2 ≤ m0 * n + n1

Rule argument

spec: nat-mult inst: 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] flags: current, []

Additional information

Used simplifier rules: mult-c: ⊦ m * n = n * m;%ler: ⊦ m ≤ m; from specification nat%zero: ⊦ 0 < m ↔ m ≠ 0; from specification nat-basic1%add-base: ⊦ n + 0 = n; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%anti: ⊦ m ≤ n → (n ≤ m ↔ m = n); from specification nat%le-add: ⊦ m + m0 ≤ m ↔ m0 = 0; from specification nat Used terms: n, m, m0, m0, n0, 0