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Proof step 218 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, bits0 + bitsand(bits, int2bits(j))) = lastn(8, bitsand(bits, int2bits(j)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 219:
# bits = 8, int2bits(i) = bits0 + bits, # int2bits(j) < # bits0 + # bitsand(bits, int2bits(j)), 8 ≤ # bits0 + # bitsand(bits, int2bits(j)), j < 0, # int2bits(j) < 8 ⊦ lastn(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 ≠ 0;% ⊦ m < n ∧ n ≤ n0 → m < n0; (forward)%0 < n ⊦ (* n) ≤ m → m ≠ 0; (forward)% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ #(bits + bits0) = # bits + # bits0;