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Proof step 6 from proof le-01 (previous step)

n0 < n * (n0 + n) / n, (n0 + n) / n * n ≤ n0 + n, (n0 + n) / n ≤ n0 / n ⊦ (n0 + n) / n = 0, n = 0

User applied rule insert lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: Divdef seq: ⊦ n ≠ 0 → m / n * n ≤ m ∧ m < (m / n)+1 * n subst: [n, m] → [n, n0]

Additional information

Used simplifier rules: ⊦ m +1 = m + 1; (s)% ⊦ m * (n + 1) = m * n + m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m < n → n ≠ 0; (s)% ⊦ m * n < m1 * n + n ↔ m < m1 + 1 ∧ n ≠ 0; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ ¬ n < n; (s)% ⊦ m0 ≠ 0 → (m0 * m < m0 * n ↔ m < n); (s)% ⊦ m ≤ n → (n < m ↔ false); (s) Used terms: n, n0