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Proof step 141 from proof byte-law-and-one (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward)% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ int2bits(j).first ↔ j < 0;% ⊦ fillsgn(n, int2bits(j)).first ↔ int2bits(j).first;