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Proof step 15 from proof fits-law-bor (previous step)

b2i(by1) ∈byte ⊦ 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)) = bcompl(b2i(by1)) ∧ bcompl(b2i(by2)) ∈byte

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
⊦ 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(by2)) ∈byte

Rule argument

Additional information

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