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Proof step 203 from proof byte-law-and-one (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → m ≤ n;% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ int2bits(j).first ↔ j < 0;% ⊦ int2bits(j) ≠ [];