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Proof step 66 from proof byte-law-and-one (previous step)

¬(int2bits(i).first ∧ # int2bits(i) ≤ # int2bits(j) ∧ int2bits(i) ≠ []), 6 = # fillsgn(6, int2bits(i)), # int2bits(j) < # bitsand(int2bits(j), int2bits(i)), # int2bits(i) < 8, 8 < # bitsand(int2bits(j), int2bits(i)), # int2bits(j) < 8, # int2bits(i) < # int2bits(j), # int2bits(i) < 7, # int2bits(i) + 1 < # bitsand(int2bits(j), int2bits(i)), # int2bits(i) < # bitsand(int2bits(j), int2bits(i)), # int2bits(j) + 1 < # bitsand(int2bits(j), int2bits(i)) ⊦ lastn(8, bitsand(int2bits(j), int2bits(i))) = lastn(8, fillsgn(8, bitsand((¬ 0 ≤ i)' + (¬ 0 ≤ i)' + fillsgn(6, int2bits(i)), int2bits(j)))), 8 = # fillsgn(8, bitsand(int2bits(j), int2bits(i)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 67:
6 = # fillsgn(6, int2bits(i)), # int2bits(j) < # bitsand(int2bits(j), int2bits(i)), # int2bits(i) < 8, 8 < # bitsand(int2bits(j), int2bits(i)), # int2bits(j) < 8, # int2bits(i) < # int2bits(j), # int2bits(i) < 7, # int2bits(i) + 1 < # bitsand(int2bits(j), int2bits(i)), # int2bits(i) < # bitsand(int2bits(j), int2bits(i)), # int2bits(j) + 1 < # bitsand(int2bits(j), int2bits(i)), 0 ≤ i ⊦ 8 = # fillsgn(8, bitsand(int2bits(j), int2bits(i))), lastn(8, bitsand(int2bits(j), int2bits(i))) = lastn(8, fillsgn(8, bitsand(false ' + false ' + fillsgn(6, int2bits(i)), int2bits(j))))

Rule argument

Additional information

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