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Proof step 2 from proof div-minus255-256 (previous step)

i < 0 ⊦ i % 256 = 0, (i - 255) / 256 = i / 256 - 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
i < 0 ⊦ i % 256 = 0, i / 256 = (i - 255) / 256 + 1

Rule argument

Additional information

Used simplifier rules: forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs% ⊦ i - j = i0 ↔ i = i0 + j; (s)