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Proof step 5 from proof mod-grz-01 (previous step)

i < 0, 0 < i % j, i + 1 < i % j, abs(j) = 1 ∨ abs(j) + i ≤ 0 ⊦ j = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

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

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: mod-pos-not-neg: ⊦ 0 < i1 ∧ 0 < i % i1 → ¬ i < 0;%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%lst-01: ⊦ ¬ i < j → (i ≠ j ↔ j < i); from specification int-basic1 Used terms: i, j, i0, j0