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Proof step 14 from proof byte-law-add-one (previous step)

i < 0 ⊦ i % 256 < 0, i2b(i + j) = i2b(b2i(i2b(i)) + j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
i % 256 = 0, i < 0 ⊦ i2b(i + j) = i2b(b2i(i2b(i)) + j)

Rule argument

Additional information

Used simplifier rules: 0 < j ⊦ i % (* j) < 0 ↔ i < 0 ∧ i % j ≠ 0;