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Proof step 2 from proof byte-law-mult-one (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
i < 0 ⊦ i2b(i * j) = i2b(b2i(i2b(i)) * j), i % 256 = 0

Rule argument

Additional information

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