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Proof step 6 from proof byte-law-add-one (previous step)

i < 0 ⊦ i2b(i + j) = i2b(b2i(i2b(i)) + j)

User applied rule cut formula

And obtained the following new goals

Proof step 7:
i % 256 < 0, i < 0 ⊦ i2b(i + j) = i2b(b2i(i2b(i)) + j)
Proof step 14:
i < 0 ⊦ i % 256 < 0, i2b(i + j) = i2b(b2i(i2b(i)) + j)

Rule argument

fma: i % 256 < 0

Additional information

Used formulas: i % 256 < 0