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Proof step 271 from proof byte-law-and-one (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 272:
# bits = 8, int2bits(i) = bits0 + bits, # bits1 = 8, int2bits(j) = bits2 + bits1, 8 ≤ # bitsand(bits1, bits), 8 ≤ # bitsand(bits0, bits2) + # bitsand(bits1, bits) ⊦ bits = [], bits1 = [], bits0 = [], bits2 = [], bits2 + bits1 = [], lastn(8, bitsand(bits1, bits)) = lastn(8, fillsgn(8, bitsand(bits2 + bits1, bits))), # bitsand(bits0, bits2) + # bitsand(bits1, bits) = 0

Rule argument

Additional information

Used simplifier rules: ⊦ # bits = 0 ↔ bits = [];%0 < n ⊦ (* n) ≤ m → m ≠ 0; (forward)% ⊦ m + n = n + m;% ⊦ bits ≠ [] ∧ bits0 ≠ [] → bitsand(bits, bits0) ≠ [];