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Proof step 3 from proof byte-law-sub-one-01 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ i2b(~ i) = i2b(~ b2i(i2b(i)))
Proof step 5:
⊦ i2b(i - j) = i2b(~ b2i(i2b(b2i(i2b(j)) + ~ i)))

Rule argument

name: byte-law-neg rotate: F subst: [i] → [b2i(i2b(j)) + ~ i]

Additional information

Used terms: b2i(i2b(j)) + ~ i