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Proof step 233 from proof byte-law-and-one (previous step)

¬(# int2bits(j) ≤ # bits ∧ int2bits(j) ≠ [] ∧ ¬ int2bits(j).first), # bits = 8, int2bits(i) = bits0 + bits, 0 ≤ j, 0 < j, j < 128, 64 ≤ j ⊦ lastn(8, fillsgn(8, bitsand(bits0 + bits, int2bits(j)))) = lastn(8, fillsgn(8, bitsand(bits, int2bits(j)))), bitsand(bits, int2bits(j)) = [], bits0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ # bits = 0 ↔ bits = [];% ⊦ m < n → n ≠ 0; (forward)% ⊦ m < n ∧ m = n0 → n0 < n; (forward)% ⊦ i < j → ¬ j < i;% ⊦ int2bits(j).first ↔ j < 0;% ⊦ int2bits(j) ≠ [];% ⊦ ¬ m ≤ n ↔ n < m;