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Proof step 5 from proof div2-nls-02 (previous step)

(i + j) / 2 < i, i ≤ j, (i + j) / 2 < j, i + j < 0 ⊦

User applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
⊦ 0 < j ∧ i < 0 → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ abs(j0) < j ∧ i0 ≤ 0)
Proof step 7:
i + j = i0 * 2 + j0 ∧ j0 ≤ 0 ∧ abs(j0) < 2 ∧ i0 ≤ 0 ∧ 0 < 2, i0 < i, i ≤ j, i0 < j, i + j < 0 ⊦

Rule argument

spec: int-div inst: name: div-elim-negpos seq: ⊦ 0 < j ∧ i < 0 → ( i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ abs(j0) < j ∧ i0 ≤ 0) subst: [i, j, i0, j0] → [i + j, 2, i0, j0] flags: rewrite, current

Additional information

Used simplifier rules: zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs Used terms: i + j, 2, i0, j0