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Proof step 20 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 21:
0 ≤ i % 256 ⊦ i2b(i * j) = i2b(b2i(i2b(i)) * j)

Rule argument

Additional information

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