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Proof step 167 from proof byte-law-and-one (previous step)

# int2bits(i) ≤ # bits1 ∧ int2bits(i) ≠ [] ∧ int2bits(i).first, # bits2 + # bits1 = 8, int2bits(j) = (bits0 + bits2) + bits1, # bits1 = # int2bits(i), 7 = # fillsgn(7, int2bits(i)), 8 ≤ # bitsand(bits2 + bits1, fillsgn(8, int2bits(i))), 8 ≤ # bits0 + # bitsand(bits2 + bits1, fillsgn(8, int2bits(i))), # bits1 < 8, i < 0, # bits1 < # bitsand(bits2 + bits1, fillsgn(8, int2bits(i))), # bits1 < # bits0 + # bitsand(bits2 + bits1, fillsgn(8, int2bits(i))) ⊦ bitsand(bits2 + bits1, fillsgn(8, int2bits(i))) = [], lastn(8, fillsgn(8, (bits0 + bits2) + bitsand(bits1, int2bits(i)))) = lastn(8, bitsand(bits2 + bits1, fillsgn(8, int2bits(i)))), bitsand(bits2 + bits1, fillsgn(8, int2bits(i))) = [], bits2 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 168:
7 = # fillsgn(7, int2bits(i)), # bits1 = # int2bits(i), int2bits(j) = (bits0 + bits2) + bits1, # bits2 + # bits1 = 8, # bits1 < 8, # bits1 < # bits0 + # bitsand(bits2 + bits1, fillsgn(8, int2bits(i))), # bits1 < # bitsand(bits2 + bits1, fillsgn(8, int2bits(i))), i < 0, 8 ≤ # bits0 + # bitsand(bits2 + bits1, fillsgn(8, int2bits(i))), 8 ≤ # bitsand(bits2 + bits1, fillsgn(8, int2bits(i))) ⊦ bits2 = [], bitsand(bits2 + bits1, fillsgn(8, int2bits(i))) = [], lastn(8, fillsgn(8, (bits0 + bits2) + bitsand(bits1, int2bits(i)))) = lastn(8, bitsand(bits2 + bits1, fillsgn(8, int2bits(i))))

Rule argument

Additional information

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