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Proof step 10 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))) ∧ b2i(i2b(band(bcompl(b2i(by1)), b2i(by2)))) = band(bcompl(b2i(by1)), b2i(by2))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ i ∈byte ∧ j ∈byte → b2i(i2b(band(i, j))) = band(i, j)
Proof step 12:
¬(b2i(by1) ∈byte ∧ bcompl(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))))), 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))
Proof step 17:
b2i(by1) ∈byte ∧ bcompl(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))))), 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: band-bib rotate: F subst: [j, i] → [bcompl(b2i(by2)), b2i(by1)]

Additional information

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