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Proof step 197 from proof byte-law-and-one (previous step)

int2bits(i) = bits0 + bits, # bits = 8 ⊦ 8 ≤ # int2bits(j), bitsand(bits, int2bits(j)) = [], lastn(8, fillsgn(8, bitsand(bits0 + bits, int2bits(j)))) = lastn(8, fillsgn(8, bitsand(bits, int2bits(j))))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 198:
# bits = 8, int2bits(i) = bits0 + bits, # int2bits(j) < 8 ⊦ lastn(8, fillsgn(8, bitsand(bits0 + bits, int2bits(j)))) = lastn(8, fillsgn(8, bitsand(bits, int2bits(j)))), bitsand(bits, int2bits(j)) = []

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (forward)% ⊦ m < n ∧ n = n0 → m < n0; (forward)% ⊦ ¬ m ≤ n ↔ n < m;