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Proof step 227 from proof byte-law-and-one (previous step)

bits2 = [], int2bits(j) = bits2 + bits1, # bits1 = 8, int2bits(i) = bits0 + bits, # bits = 8 ⊦ bits0 = [], bitsand(bits, bits2 + bits1) = [], lastn(8, fillsgn(8, bitsand(bits0 + bits, bits2 + bits1))) = lastn(8, fillsgn(8, bitsand(bits, bits2 + bits1)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 228:
# bits = 8, int2bits(i) = bits0 + bits, # int2bits(j) = 8 ⊦ lastn(8, fillsgn(8, bitsand(bits0 + bits, int2bits(j)))) = lastn(8, fillsgn(8, bitsand(bits, int2bits(j)))), bitsand(bits, int2bits(j)) = [], bits0 = []

Rule argument

Additional information

Used simplifier rules: ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ [] + bits = bits;