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Proof step 6 from proof fits-law-bor (previous step)

⊦ bcompl(band(bcompl(b2i(by1)), bcompl(b2i(by2)))) = b2i(i2b(bcompl(band(bcompl(b2i(by1)), bcompl(b2i(by2)))))), band(bcompl(b2i(by1)), bcompl(b2i(by2))) ∈byte

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ i ∈byte ↔ b2i(i2b(i)) = i
Proof step 8:
⊦ bcompl(band(bcompl(b2i(by1)), bcompl(b2i(by2)))) = b2i(i2b(bcompl(band(bcompl(b2i(by1)), bcompl(b2i(by2)))))), b2i(i2b(band(bcompl(b2i(by1)), bcompl(b2i(by2))))) = band(bcompl(b2i(by1)), bcompl(b2i(by2)))

Rule argument

name: inbyte-bib rotate: F subst: [i] → [band(bcompl(b2i(by1)), bcompl(b2i(by2)))]

Additional information

Used terms: band(bcompl(b2i(by1)), bcompl(b2i(by2)))