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Proof step 212 from proof byte-law-and-one (previous step)

int2bits(j).first ∧ # int2bits(j) ≤ # bits ∧ int2bits(j) ≠ [], # bits = 8, int2bits(i) = bits0 + bits, # int2bits(j) < # bits, # int2bits(j) < 8, j < 0, 8 ≤ # bits ⊦ lastn(8, fillsgn(8, bits0 + bitsand(bits, int2bits(j)))) = lastn(8, bitsand(bits, int2bits(j))), bitsand(bits, int2bits(j)) = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 213:
int2bits(i) = bits0 + bits, # bits = 8, # int2bits(j) < 8, j < 0 ⊦ bitsand(bits, int2bits(j)) = [], lastn(8, fillsgn(8, bits0 + bitsand(bits, int2bits(j)))) = lastn(8, bitsand(bits, int2bits(j)))

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward)% ⊦ m < n → m ≤ n;% ⊦ j < 0 → int2bits(j).first;% ⊦ int2bits(j) ≠ [];