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Proof step 12 from proof fits-law-bxor (previous step)

¬(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))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
⊦ 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))))), bcompl(b2i(by2)) ∈byte, 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

Additional information

Used simplifier rules: byte2int-in: ⊦ b2i(by) ∈byte;