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Proof step 2 from proof div-add1-posneg-neg (previous step)

i % j = 0, i + i0 < 0, i0 < 0, 0 < j, 0 ≤ i ⊦ (i + i0) / j = i / j + i0 / j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
i % j = 0, i + i0 < 0, i0 < 0, 0 < j, 0 ≤ i ⊦ (i + i0) / j = i / j + i0 / j

Rule argument

Additional information

Used simplifier rules: add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i < 0 ∧ 0 ≤ j → i < j; (s)% ⊦ i + j < j ↔ i < 0; (s)%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs