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Proof step 242 from proof byte-law-and-one (previous step)

8 ≤ # bitsand(bits, int2bits(j)), # bits = 8, int2bits(i) = bits0 + bits, # int2bits(j) ≤ 8, ~128 ≤ j, j < ~64, j < 0 ⊦ lastn(8, fillsgn(8, bits0 + bitsand(bits, int2bits(j)))) = lastn(8, bitsand(bits, int2bits(j))), bitsand(bits, int2bits(j)) = [], bits0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 243:
int2bits(i) = bits0 + bits, # bits = 8, # int2bits(j) ≤ # bitsand(bits, int2bits(j)), j < 0, j < ~64, ~128 ≤ j, # int2bits(j) ≤ 8, 8 ≤ # bitsand(bits, int2bits(j)) ⊦ bits0 = [], bitsand(bits, int2bits(j)) = [], lastn(8, fillsgn(8, bits0 + bitsand(bits, int2bits(j)))) = lastn(8, bitsand(bits, int2bits(j)))

Rule argument

Additional information

Used simplifier rules: ⊦ # bits = 0 ↔ bits = [];%0 < n ⊦ (* n) ≤ m → m ≠ 0; (forward)% ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward)