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Proof step 3 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)))))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ i ∈byte ∧ j ∈byte → b2i(i2b(bor(i, j))) = bor(i, j)
Proof step 5:
¬(band(b2i(by1), bcompl(b2i(by2))) ∈byte ∧ band(bcompl(b2i(by1)), b2i(by2)) ∈byte) ⊦ 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)))))

Rule argument

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

Additional information

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