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Proof step 2 from proof zero-06 (previous step)

i < 0, j < 0, i / j < 0 ⊦ j = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
⊦ j < 0 ∧ i < 0 → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ j < j0)

Rule argument

spec: int-div inst: name: div-elim3 seq: ⊦ j < 0 ∧ i < 0 → ( i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ j < j0) subst: [i, j, i0, j0] → [i, j, i0, j0] flags: rewrite, current

Additional information

Used simplifier rules: modzero-04: ⊦ i < 0 ∧ j < j0 ∧ j < 0 → ¬ i * j + j0 < 0; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs Used terms: i, j, i0, j0