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Proof step 83 from proof byte-law-and-one (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → m ≤ n;% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ ¬ m < n ↔ n ≤ m;% ⊦ m < n → ¬ n < m;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;%n < n0 ⊦ m + (* n) < (* n0) ↔ m < n0 - n;% ⊦ m +1 = m + 1;% ⊦ # bits = 0 ↔ bits = [];% ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward)% ⊦ m < n → n ≠ 0; (forward)% ⊦ m < n ∧ m = n0 → n0 < n; (forward)% ⊦ n < n0 ∧ n0 < n1 → n +1 < n1; (forward)% ⊦ int2bits(j) ≠ [];% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ int2bits(j).first ↔ j < 0;