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Proof step 5 from proof div-minus-onediff-01 (previous step)

j ≠ 0, i + 1 = i0 + j, ∃ j0. i = j0 * j, j + i ≤ 0, i < j, i + 1 < j, 0 < j, i < 0 ⊦ j = 1, i0 / j = i / j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
j0 * j + 1 = i0 + j, i0 < j, i0 < 0, 0 < j, j + j0 * j ≤ 0, j0 < 0, j0 < j, j0 + 1 < j, i0 + 1 < j ⊦ j = 0, j = 1, i0 / j = j0

Rule argument

Additional information

Used simplifier rules: div-mult-03: j ≠ 0 ⊦ (i * j) / j = i;%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%ls: ⊦ i + j < j ↔ i < 0; 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%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%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