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Proof step 4 from proof div-minus-onediff (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
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 = ((i0 * j + j0) + 1 - j) / j + 1

Rule argument

Additional information

Used simplifier rules: forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%lsone: ⊦ i < 1 ↔ i ≤ 0; from specification int-abs%ls-same: ⊦ 0 < i → (i * j < i ↔ j < 1); from specification int-mult%lst-03: ⊦ i + j < i0 ∧ ¬ j < j0 → i + j0 < i0; from specification int-basic2%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%ls-add: ⊦ 0 ≤ j1 ∧ j1 < j → (i0 * j + j1 < j ↔ i0 ≤ 0); from specification int-mult%add-less: ⊦ 0 < i → i1 < i1 + i; 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%ls-01: ⊦ i + j < i + i0 ↔ j < i0; from specification int-basic2%succ-01: ⊦ i +1 = i + 1; from specification int-basic2%forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%neg-13: ⊦ ~ i < j ↔ 0 < j + i; from specification int-abs%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs