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Proof step 24 from proof mod-ls-same-02 (previous step)

j < 0, 0 < i % j, 0 ≤ i, i < i % j ⊦ j = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 25:
⊦ j < 0 ∧ 0 ≤ i → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < abs(j) ∧ i0 ≤ 0)

Rule argument

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

Additional information

Used simplifier rules: abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs%neg-07: ⊦ i < ~ j ↔ i + j < 0; from specification int-abs%ls: ⊦ i + j < j ↔ i < 0; from specification int-basic2%zero-ls-01: ⊦ i < 0 → (i * j < 0 ↔ 0 < j); from specification int-mult%lels-04: ⊦ i ≤ j → (¬ ¬ j < i ↔ false); from specification int-abs%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 Used terms: i, j, i0, j0