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Proof step 55 from proof byte-law-and-one (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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