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Proof step 52 from proof byte-law-and-one (previous step)

8 = # fillsgn(8, int2bits(i)), # int2bits(i) < 8 ⊦ 8 ≤ # int2bits(j), lastn(8, fillsgn(8, bitsand(int2bits(i), int2bits(j)))) = lastn(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 ∧ n = n0 → m < n0; (forward)% ⊦ ¬ m ≤ n ↔ n < m;