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Proof step 11 from proof fits-law-bor (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 12:
⊦ i ∈byte ↔ b2i(i2b(i)) = i
Proof step 13:
⊦ b2i(i2b(band(bcompl(b2i(by1)), bcompl(b2i(by2))))) = band(bcompl(b2i(by1)), bcompl(b2i(by2))), bcompl(band(bcompl(b2i(by1)), bcompl(b2i(by2)))) = b2i(i2b(bcompl(band(bcompl(b2i(by1)), bcompl(b2i(by2)))))), b2i(i2b(bcompl(b2i(by1)))) = bcompl(b2i(by1)) ∧ bcompl(b2i(by2)) ∈byte

Rule argument

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

Additional information

Used terms: bcompl(b2i(by1))