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Proof step 1 from proof byte-law-mult-one (conclusion)

⊦ i2b(i * j) = i2b(b2i(i2b(i)) * j)

User applied rule cut formula

And obtained the following new goals

Proof step 2:
i % 256 < 0 ⊦ i2b(i * j) = i2b(b2i(i2b(i)) * j)
Proof step 20:
⊦ i % 256 < 0, i2b(i * j) = i2b(b2i(i2b(i)) * j)

Rule argument

fma: i % 256 < 0

Additional information

Used formulas: i % 256 < 0