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Proof step 41 from proof byte-law-and-one (previous step)

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

Additional information

Used simplifier rules: ⊦ bitsand(int2bits(i), int2bits(i0)) ≠ [];