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Proof step 209 from proof byte-law-and-one (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 210:
# bits = 8, int2bits(i) = bits0 + bits, # int2bits(j) < # bitsand(bits, int2bits(j)), # int2bits(j) < 8, j < 0, 8 ≤ # bitsand(bits, int2bits(j)) ⊦ lastn(8, fillsgn(8, bits0 + bitsand(bits, int2bits(j)))) = lastn(8, bitsand(bits, int2bits(j))), bitsand(bits, int2bits(j)) = []

Rule argument

Additional information

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