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Proof step 8 from proof div-le-same-neg (previous step)

j = i1 + 1, i0 * j + j0 < j, i0 * j + j0 < 0, i0 < j, i0 ≤ 0, j0 < j, j0 ≤ 0, (i0 * j + j0) + 1 < j, 0 < j, abs(j0) < j ⊦ 0 ≤ i1 * i0 + j0 ↔ i0 * j + j0 = 0 ∨ j = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%forward-lele: ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward) from specification int-abs%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%zero-gels-03: ⊦ i < 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%mone-06: ⊦ ~1 < i ↔ 0 ≤ i; from specification int-abs%numls-02: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; from specification int-abs%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%numls-01: ⊦ i + (* j) < i0 + (* j0) ↔ i < i0 + (j0 - j); from specification int-abs%succ-02: ⊦ i < j + 1 ↔ i ≤ j; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%numadd-01: ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; from specification int-abs