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Proof step 3 from proof ls-02 (previous step)

m / n < m0 / n, m0 ≤ m ⊦ n = 0

User applied rule insert lemma

And obtained the following new goals

Proof step 4:
⊦ n ≠ 0 → m / n * n ≤ m ∧ m < (m / n)+1 * n
Proof step 5:
m / n < m0 / n, m0 ≤ m, n ≠ 0 → m0 / n * n ≤ m0 ∧ m0 < (m0 / n)+1 * n ⊦ n = 0

Rule argument

spec: <toplevel> name: Divdef seq: ⊦ n ≠ 0 → m / n * n ≤ m ∧ m < (m / n)+1 * n subst: [n, m] → [n, m0]

Additional information

Used terms: n, m0