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Proof step 2 from proof div-minus-onediff-01 (previous step)

i % j = 0, j0 = j - 1, i < 0, 0 < j ⊦ (i - j0) / j = i / j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
i % j = 0, 0 < j, i < 0 ⊦ (i + 1 - j) / j = i / j

Rule argument

Additional information

Used simplifier rules: 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% ⊦ i - (j - j0) = i + j0 - j; (s)