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Proof step 72 from proof byte-law-and-one (previous step)

¬(int2bits(j).first ∧ # int2bits(j) ≤ # int2bits(i) ∧ int2bits(j) ≠ []), 8 = # fillsgn(8, int2bits(i)), # int2bits(i) < # bitsand(int2bits(i), int2bits(j)), # int2bits(i) < 8, 8 < # bitsand(int2bits(i), int2bits(j)), # int2bits(j) + 1 < # bitsand(int2bits(i), int2bits(j)), # int2bits(j) < # bitsand(int2bits(i), int2bits(j)), # int2bits(i) + 1 < # bitsand(int2bits(i), int2bits(j)), # int2bits(j) ≤ # int2bits(i), # int2bits(j) < 8 ⊦ lastn(8, bitsand(int2bits(i), int2bits(j))) = lastn(8, fillsgn(8, bitsand(fillsgn(8, int2bits(i)), int2bits(j)))), 8 = # fillsgn(8, bitsand(int2bits(i), int2bits(j))), bitsand(fillsgn(8, int2bits(i)), int2bits(j)) = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 73:
8 = # fillsgn(8, int2bits(i)), # int2bits(i) < # bitsand(int2bits(i), int2bits(j)), # int2bits(i) < 8, 8 < # bitsand(int2bits(i), int2bits(j)), # int2bits(j) + 1 < # bitsand(int2bits(i), int2bits(j)), # int2bits(j) < # bitsand(int2bits(i), int2bits(j)), # int2bits(i) + 1 < # bitsand(int2bits(i), int2bits(j)), # int2bits(j) ≤ # int2bits(i), # int2bits(j) < 8, 0 ≤ j ⊦ lastn(8, bitsand(int2bits(i), int2bits(j))) = lastn(8, fillsgn(8, bitsand(fillsgn(8, int2bits(i)), int2bits(j)))), 8 = # fillsgn(8, bitsand(int2bits(i), int2bits(j))), bitsand(fillsgn(8, int2bits(i)), int2bits(j)) = []

Rule argument

Additional information

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