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Proof step 2 from proof byte-law-sub-one-01 (previous step)

i - b2i(i2b(j)) = ~(b2i(i2b(j)) + ~ i) ⊦ i2b(i - j) = i2b(i - b2i(i2b(j)))

User applied rule insert equation

And obtained the following new goals

Proof step 3:
⊦ i2b(i - j) = i2b(~(b2i(i2b(j)) + ~ i))

Rule argument

pos: left 1 flags: discard

Additional information