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Proof step 91 from proof byte-law-and-one (previous step)

fillsgn(8, int2bits(i)) ≠ [] ∧ bits0 + bits ≠ [], 8 = # fillsgn(8, int2bits(i)), # bits = 8, int2bits(j) = bits0 + bits, # int2bits(i) < 8 ⊦ bitsand(bits0 + bits, fillsgn(8, int2bits(i))) = [], lastn(8, fillsgn(8, bitsand(int2bits(i), bits0 + bits))) = lastn(8, fillsgn(8, bitsand(bits0 + bits, fillsgn(8, int2bits(i)))))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ bits ≠ [] ∧ bits0 ≠ [] → bitsand(bits, bits0) ≠ [];