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Proof step 259 from proof byte-law-and-one (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 260:
# 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(bits0, bits2) + bitsand(bits, bits1)) = lastn(8, fillsgn(8, bitsand(bits2 + bits1, bits))), # bitsand(bits0, bits2) + # bitsand(bits, bits1) = 0

Rule argument

Additional information

Used simplifier rules: 0 < n ⊦ (* n) ≤ m → m ≠ 0; (forward)% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ #(bits + bits0) = # bits + # bits0;