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Proof step 22 from proof byte-law-mult-one (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
i % 256 = 0, i < 0 ⊦ i2b(i * j) = i2b(b2i(i2b(i)) * j)

Rule argument

Additional information

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