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Proof step 268 from proof byte-law-and-one (previous step)

8 ≤ # bitsand(bits, bits1), # bits = 8, int2bits(i) = bits0 + bits, # bits1 = 8, int2bits(j) = bits2 + bits1, 8 ≤ # bitsand(bits0, bits2) + # bitsand(bits, bits1) ⊦ bits0 = [], bits2 = [], bits = [], bits2 + bits1 = [], lastn(8, bitsand(bits, bits1)) = lastn(8, fillsgn(8, bitsand(bits2 + bits1, bits))), # bitsand(bits0, bits2) + # bitsand(bits, bits1) = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 269:
int2bits(j) = bits2 + bits1, # bits1 = 8, int2bits(i) = bits0 + bits, # bits = 8, 8 ≤ # bitsand(bits0, bits2) + # bitsand(bits, bits1), 8 ≤ # bitsand(bits, bits1) ⊦ # bitsand(bits0, bits2) + # bitsand(bits, bits1) = 0, lastn(8, bitsand(bits, bits1)) = lastn(8, fillsgn(8, bitsand(bits2 + bits1, bits))), bits2 + bits1 = [], bits = [], bits2 = [], bits0 = [], bitsand(bits, bits1) = []

Rule argument

Additional information

Used simplifier rules: ⊦ # bits = 0 ↔ bits = [];%0 < n ⊦ (* n) ≤ m → m ≠ 0; (forward)