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Proof step 9 from proof div-minus-onediff (previous step)

(j * (i0 - 1)) % j = 0 ∧ j * (i0 - 1) < 0 ∧ j0 + 1 < 0 ∧ 0 < j, j0 < j, j0 < 0, i0 * j + j0 < j, i0 * j + j0 < 0, i0 < j + j0, i0 < j, i0 ≤ 0, (i0 * j + j0) + 1 < j, 0 < j + j0, j0 + 1 < j, 1 < j, 0 < j ⊦ j0 = 0, i0 = ((j * (i0 - 1)) / j + (j0 + 1) / j) + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: div-mult-03: j ≠ 0 ⊦ (i * j) / j = i;%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%numls-03: ⊦ i + (* j) < (* j0) ↔ i < j0 - j; from specification int-abs%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%lsone: ⊦ i < 1 ↔ i ≤ 0; from specification int-abs%sub-ls: ⊦ i - j < i0 ↔ i < j + i0; from specification int-abs%zero-ls: ⊦ 0 < i → (i * j < 0 ↔ j < 0); from specification int-mult%eq: ⊦ i = i + j ↔ j = 0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%Sub-def: ⊦ i + (j - i) = j; from specification int-abs%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%zero-left: ⊦ 0 * i = 0; from specification int-mult