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Proof step 4 from proof mod-lsz (previous step)

i < j, i < 0, i % j + 1 < j, j + i % j < 0, 0 < j, i % j < ~1, i % j < 0, i + 1 < j ⊦

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 5:
⊦ 0 < j ∧ i < 0 → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ abs(j0) < j)

Rule argument

spec: int-div inst: name: div-elim2 seq: ⊦ 0 < j ∧ i < 0 → ( i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ abs(j0) < j) subst: [i, j, i0, j0] → [i, j, i0, j0] flags: rewrite, current

Additional information

Used simplifier rules: zero-gels-03: ⊦ i < 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs%neg-13: ⊦ ~ i < j ↔ 0 < j + i; from specification int-abs%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs Used terms: i, j, i0, j0