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Proof step 38 from proof byte-law-and-one (previous step)

bitsand(lastn(8, fillsgn(8, int2bits(i))), int2bits(j)) ≠ [] ⊦ lastn(8, fillsgn(8, sgn(bitsand(int2bits(i), int2bits(j))))) = lastn(8, fillsgn(8, bitsand(lastn(8, fillsgn(8, int2bits(i))), int2bits(j))))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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