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Proof step 150 from proof byte-law-and-one (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 151:
int2bits(j) = bits0 + bits, # bits = 8, 8 = # fillsgn(8, int2bits(i)), # bitsand(bits, fillsgn(8, int2bits(i))) < 8, i < 0, # int2bits(i) < 8, 8 ≤ # bits0 + # bitsand(bits, fillsgn(8, int2bits(i))), # int2bits(i) < # bits0 + # bitsand(bits, fillsgn(8, int2bits(i))) ⊦ fillsgn(8, int2bits(i)) = [], bits0 + bits = [], lastn(8, fillsgn(8, bitsand(int2bits(i), bits0 + bits))) = lastn(8, bits0 + bitsand(bits, fillsgn(8, int2bits(i)))), bits0 = []

Rule argument

Additional information

Used simplifier rules: ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ # bits = 0 ↔ bits = [];% ⊦ m < m + n ↔ n ≠ 0;% ⊦ m < n ∧ n = n0 → m < n0; (forward)% ⊦ m < n ∧ n ≤ n0 → m < n0; (forward)% ⊦ ¬ m ≤ n ↔ n < m;