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Proof step 2 from proof byte-add-256-01 (previous step)

⊦ i2b(~(i + 256)) = i2b(~ i)

User applied rule cut formula

And obtained the following new goals

Proof step 3:
~(i + 256) = ~(1 * 256 + i) ⊦ i2b(~(i + 256)) = i2b(~ i)

Rule argument

fma: ~(i + 256) = ~(1 * 256 + i)

Additional information

Used simplifier rules: ⊦ i + j = j + i;% ⊦ ~ i = ~ j ↔ i = j; Used formulas: ~(i + 256) = ~(1 * 256 + i)