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Proof step 5 from proof fits-law-bor (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
⊦ bcompl(band(bcompl(b2i(by1)), bcompl(b2i(by2)))) = b2i(i2b(bcompl(band(bcompl(b2i(by1)), bcompl(b2i(by2)))))), band(bcompl(b2i(by1)), bcompl(b2i(by2))) ∈byte

Proof step 5 from proof fits-law-bor (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information