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Proof step 247 from proof byte-law-and-one (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 248:
int2bits(j) = bits2 + bits1, # bits1 = 8, int2bits(i) = bits0 + bits, # bits = 8 ⊦ bits2 = [], bits0 = [], lastn(8, fillsgn(8, bitsand(bits0 + bits, bits2 + bits1))) = lastn(8, fillsgn(8, bitsand(bits2 + bits1, bits))), bits2 + bits1 = [], bits = []

Rule argument

Additional information

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