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Proof step 239 from proof byte-law-and-one (previous step)

¬ 8 ≤ # bitsand(bits, int2bits(j)), # bits = 8, int2bits(i) = bits0 + bits, # int2bits(j) ≤ 8, ~128 ≤ j, j < ~64, j < 0 ⊦ lastn(8, fillsgn(8, bits0 + bitsand(bits, int2bits(j)))) = lastn(8, fillsgn(8, bitsand(bits, int2bits(j)))), bitsand(bits, int2bits(j)) = [], bits0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 240:
int2bits(i) = bits0 + bits, # bits = 8, # bitsand(bits, int2bits(j)) < 8, j < 0, j < ~64, ~128 ≤ j, # int2bits(j) ≤ 8 ⊦ lastn(8, fillsgn(8, bits0 + bitsand(bits, int2bits(j)))) = lastn(8, bitsand(bits, int2bits(j)).first ' + fillsgn(7, bitsand(bits, int2bits(j)))), bits0 = [], bitsand(bits, int2bits(j)) = []

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (forward)% ⊦ m < n ∧ n = n0 → m < n0; (forward)%⊦ bits ≠ [] ∧ # bits < n → fillsgn(n, bits) = bits.first ' + fillsgn(n -1, bits);% ⊦ ¬ m ≤ n ↔ n < m;