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Proof step 18 from proof byte-law-add-one (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 19:
j < 0, 0 ≤ i ⊦ i % 256 < 0, i2b(i + j) = i2b(b2i(i2b(i)) + j)

Rule argument

fma: i % 256 < 0

Additional information

Used simplifier rules: 0 < j ⊦ i % (* j) < 0 ↔ i < 0 ∧ i % j ≠ 0;% ⊦ i ≤ j → (¬ ¬ j < i ↔ false); Used formulas: i % 256 < 0