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Proof step 1 from proof div-minus-onediff (conclusion)

⊦ i % j ≠ 0 ∧ j0 = j - 1 ∧ i < 0 ∧ 0 < j → (i - j0) / j = i / j - 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
i < j, i < 0, 0 < j, i + 1 < j ⊦ i / j = (i + 1 - j) / j + 1, i % j = 0

Rule argument

Additional information

Used simplifier rules: succ-01: ⊦ i +1 = i + 1; from specification int-basic2%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%sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs%sub-sub-04: ⊦ i - (j - j0) = i + j0 - j; from specification int-abs