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Proof step 17 from proof byte-law-and-one (previous step)

¬ bitsand(sgn(lastn(8, fillsgn(8, int2bits(i)))), int2bits(j)) ≠ [], lastn(8, fillsgn(8, int2bits(i))) ≠ [] ⊦ band(i, j) = bits2int(bitsand(sgn(lastn(8, fillsgn(8, int2bits(i)))), int2bits(j))), b2i(i2b(band(i, j))) = bits2int(lastn(8, fillsgn(8, int2bits(bits2int(bitsand(sgn(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: ⊦ lastn(n, bits) = [] ↔ n = 0;