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Proof step 97 from proof byte-law-and-one (previous step)

# fillsgn(8, int2bits(i)) ≤ # bits ∧ fillsgn(8, int2bits(i)) ≠ [] ∧ ¬ fillsgn(8, int2bits(i)).first, int2bits(j) = bits0 + bits, # bits = 8, 8 = # fillsgn(8, int2bits(i)), # int2bits(i) < 8, 0 ≤ i ⊦ fillsgn(8, int2bits(i)) = [], bits0 + bits = [], lastn(8, fillsgn(8, bitsand(int2bits(i), bits0 + bits))) = lastn(8, fillsgn(8, bitsand(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)% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ int2bits(j).first ↔ j < 0;% ⊦ fillsgn(n, int2bits(j)).first ↔ int2bits(j).first;