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Proof step 224 from proof byte-law-and-one (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 225:
int2bits(j) = bits2 + bits1, # bits1 = 8, int2bits(i) = bits0 + bits, # bits = 8 ⊦ bitsand(bits, bits2 + bits1) = [], lastn(8, fillsgn(8, bitsand(bits0 + bits, bits2 + bits1))) = lastn(8, fillsgn(8, bitsand(bits, bits2 + bits1)))

Proof step 224 from proof byte-law-and-one (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information