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Proof step 6 from proof mod-rec-neg (previous step)

i = i0 * j + j0 ∧ j0 ≤ 0 ∧ abs(j0) < j ∧ i0 ≤ 0, j0 = 0, i < j, i < 0, 0 < j, i + j < 0 ⊦ i |%| j = (i + j) |%| j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: mod-mult-01: ⊦ 0 < j → (i * j) |%| j = 0;%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%ls: ⊦ i + j < j ↔ i < 0; from specification int-basic2%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%transitivity: ⊦ i < i0 ∧ i0 < i1 → i < i1; (forward) from specification oelem[intlista]%zero-gels-03: ⊦ i < 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%zero-ls: ⊦ 0 < i → (i * j < 0 ↔ j < 0); from specification int-mult%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%lsone: ⊦ i < 1 ↔ i ≤ 0; from specification int-abs%ls-same: ⊦ 0 < i → (i * j < i ↔ j < 1); from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%ax-04: ⊦ i + 0 = i; from specification int-basic2