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Proof step 94 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, bitsand(bits0 + 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: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ int2bits(j).first ↔ j < 0;% ⊦ fillsgn(n, int2bits(j)).first ↔ int2bits(j).first;% ⊦ ¬ m ≤ n ↔ n < m;