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Proof step 103 from proof byte-law-and-one (previous step)

¬(# int2bits(i) ≤ # bits ∧ int2bits(i) ≠ [] ∧ int2bits(i).first), # bits = 8, int2bits(j) = bits0 + bits, 7 = # fillsgn(7, int2bits(i)), # int2bits(i) < 8, 0 ≤ i ⊦ bits0 + bits = [], lastn(8, fillsgn(8, bitsand(bits0 + bits, int2bits(i)))) = lastn(8, fillsgn(8, bitsand(bits, false ' + fillsgn(7, int2bits(i)))))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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