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Proof step 10 from proof fits-law-bor (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
⊦ 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

Proof step 10 from proof fits-law-bor (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information