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Proof step 4 from proof mod-sub-02 (previous step)

1 < i, 0 < j, j < i, i < (i0 + 1) * j, 0 < i, 1 < (i0 + 1) * j, 0 ≤ i, 0 ≤ i0, i0 * j ≤ i ⊦ i = i % j + i0 * j, i0 = i / j

User applied rule insert lemma

And obtained the following new goals

Proof step 5:
⊦ 0 ≤ i0 ∧ 0 ≤ i ∧ 0 < j ∧ i0 * j ≤ i ∧ i < (i0 + 1) * j → i0 = i / j

Rule argument

spec: <toplevel> name: is-div seq: ⊦ 0 ≤ i0 ∧ 0 ≤ i ∧ 0 < j ∧ i0 * j ≤ i ∧ i < (i0 + 1) * j → i0 = i / j subst: [i0, i, j] → [i0, i, j]

Additional information

Used simplifier rules: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s) Used terms: i0, i, j