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Proof step 5 from proof div-add1-posneg-pos-02 (previous step)

¬(1 < j ∧ i0 < 0 ∧ i0 % j ≠ 0), ¬(1 < j ∧ i0 < 0 ∧ i0 % j = 0) ⊦ i % j = 0 ∧ 0 ≤ i + i0 ∧ 0 ≤ i ∧ i0 < 0 ∧ 0 < j → (i + i0) / j = i / j + (i0 - (j - 1)) / j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
i % j = 0, i0 + 1 < j, 0 < j, 0 ≤ i + i0, 0 ≤ i, i0 < 0, i0 < j ⊦ (i + i0) / j = i / j + (i0 + 1 - j) / j, 1 < j ∧ i0 % j = 0, 1 < j ∧ i0 % j ≠ 0

Rule argument

Additional information

Used simplifier rules: succ-01: ⊦ i +1 = i + 1; from specification int-basic2%forward-le-01: ⊦ 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%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%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%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%sub-sub-04: ⊦ i - (j - j0) = i + j0 - j; from specification int-abs