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Proof step 5 from proof div-minus-onediff (previous step)

i = i0 * j + j0, i < 0, 0 < j, abs(j0) < j, j0 ≤ 0, i0 ≤ 0 ⊦ i0 = (i + 1 - j) / j + 1, j0 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
0 < j, i0 * j + j0 < 0, i0 < j, i0 < j + j0, j0 < 0, 0 < j + j0, i0 ≤ 0 ⊦ i0 = ((i0 * j + j0) + 1 - j) / j + 1, j0 = 0

Rule argument

Additional information

Used simplifier rules: 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%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs% ⊦ ~ i < j ↔ 0 < j + i; (s)% ⊦ i < 0 → abs(i) = ~ i; (s)% ⊦ i ≠ j → (i ≤ j ↔ i < j); (s)