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Proof step 8 from proof fits-law-bxor (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ i ∈byte ↔ b2i(i2b(i)) = i
Proof step 10:
⊦ bor(band(b2i(by1), bcompl(b2i(by2))), band(bcompl(b2i(by1)), b2i(by2))) = b2i(i2b(bor(band(b2i(by1), bcompl(b2i(by2))), band(bcompl(b2i(by1)), b2i(by2))))), b2i(i2b(band(b2i(by1), bcompl(b2i(by2))))) = band(b2i(by1), bcompl(b2i(by2))) ∧ b2i(i2b(band(bcompl(b2i(by1)), b2i(by2)))) = band(bcompl(b2i(by1)), b2i(by2))

Rule argument

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

Additional information

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