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Proof step 133 from proof byte-law-and-one (previous step)

bits ≠ [] ∧ fillsgn(8, int2bits(i)) ≠ [], 8 = # fillsgn(8, bitsand(fillsgn(8, int2bits(i)), bits)), 8 = # fillsgn(8, bitsand(bits, int2bits(i))), 7 = # fillsgn(7, int2bits(i)), int2bits(j) = bits0 + bits, # bits = 8, 0 ≤ i, # int2bits(i) < 8 ⊦ fillsgn(8, bitsand(bits, int2bits(i))) = fillsgn(8, bitsand(fillsgn(8, int2bits(i)), bits)), bits0 + bits = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 134:
# bits = 8, int2bits(j) = bits0 + bits, 7 = # fillsgn(7, int2bits(i)), 8 = # fillsgn(8, bitsand(bits, int2bits(i))), 8 = # fillsgn(8, bitsand(fillsgn(8, int2bits(i)), bits)), # int2bits(i) < 8, 0 ≤ i ⊦ bits = [], fillsgn(8, int2bits(i)) = [], bits0 + bits = [], fillsgn(8, bitsand(bits, int2bits(i))) = fillsgn(8, bitsand(fillsgn(8, int2bits(i)), bits))

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ∧ n = n0 → m < n0; (forward)