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Proof step 6 from proof div-add1-negpos-neg-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
i % j = 0, i + i0 < 0, i < 0, 0 < j, 0 ≤ i0 ⊦ (i + i0) / j = i / j + (i0 + j - 1) / j, 1 < j ∧ i0 % j = 0, 1 < j ∧ 0 < i0 ∧ i0 % j ≠ 0

Rule argument

Additional information

Used simplifier rules: ⊦ 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% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i + (j - i0) = i + j - i0; (s)