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Proof step 5 from proof ls-03 (previous step)

n0 < n * (n0 + n) / n, (n0 + n) / n * n ≤ n0 + n, n0 / n + 1 < (n0 + n) / n ⊦ (n0 + n) / n = 0, n = 0

User applied rule insert lemma

And obtained the following new goals

Proof step 6:
⊦ n ≠ 0 → m / n * n ≤ m ∧ m < (m / n)+1 * n
Proof step 7:
n ≠ 0 → n0 / n * n ≤ n0 ∧ n0 < (n0 / n)+1 * n, n0 < n * (n0 + n) / n, (n0 + n) / n * n ≤ n0 + n, n0 / n + 1 < (n0 + n) / n ⊦ (n0 + n) / n = 0, n = 0

Rule argument

spec: nat-div inst: name: Divdef seq: ⊦ n ≠ 0 → m / n * n ≤ m ∧ m < (m / n)+1 * n subst: [n, m] → [n, n0] flags: current, []

Additional information

Used terms: n, n0