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Proof step 69 from proof byte-law-and-one (previous step)

8 = # fillsgn(8, int2bits(i)), # int2bits(i) < # bitsand(int2bits(i), int2bits(j)), # int2bits(j) < 8, # 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(i) < # int2bits(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)) = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 70:
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)) = []

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward)% ⊦ ¬ m < n ↔ n ≤ m;