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Proof step 1 from proof div-mult-neg (conclusion)

⊦ 0 < j ∧ i ≤ 0 ∧ i0 ≤ 0 ∧ abs(i0) < j → (j * i + i0) / j = i

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
abs(i0) < j, 0 < j, i ≤ 0, i < j, i0 ≤ 0, i0 < j ⊦ (j * i + i0) / j = i

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%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