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Proof step 20 from proof div-eq-zero-01 (previous step)

j < ~1, i < 0, j < i, j ≤ i, j < 0 ⊦ i / j = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 21:
⊦ j < 0 ∧ i < 0 → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ j < j0 ∧ 0 ≤ i0)
Proof step 22:
i = i0 * j + j0 ∧ j0 ≤ 0 ∧ j < j0 ∧ 0 ≤ i0, j < ~1, i < 0, j < i, j ≤ i, j < 0 ⊦ i0 = 0

Rule argument

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