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Proof step 215 from proof byte-law-and-one (previous step)

¬ 8 ≤ #(bits0 + bitsand(bits, int2bits(j))), 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)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 216:
# bits = 8, int2bits(i) = bits0 + bits, # bits0 + # bitsand(bits, int2bits(j)) < 8, j < 0, # int2bits(j) < 8 ⊦ lastn(8, fillsgn(8, bits0 + bitsand(bits, int2bits(j)))) = lastn(8, bitsand(bits, int2bits(j))), bitsand(bits, int2bits(j)) = []

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ∧ n = n0 → m < n0; (forward)% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ #(bits + bits0) = # bits + # bits0;