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Proof step 36 from proof bib-neg-low (previous step)

¬ # int2bits(j).rest ≤ 8, i0 < 0, 256 * i0 + j < 0, j < 0, ~256 < j, 256 * i0 + j < ~128, j < ~128 ⊦ 8 = # fillsgn(8, int2bits(j).rest), b2i(i2b(bits2int(lastn(8, int2bits(i0 - 1) + fillsgn(8, int2bits(j).rest))))) = 256 + j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 37:
8 < # int2bits(j).rest, j < ~128, 256 * i0 + j < ~128, ~256 < j, j < 0, 256 * i0 + j < 0, i0 < 0 ⊦ b2i(i2b(bits2int(lastn(8, int2bits(i0 - 1) + fillsgn(8, int2bits(j).rest))))) = 256 + j, 8 = # fillsgn(8, int2bits(j).rest), int2bits(j).rest = []

Rule argument

Additional information

Used simplifier rules: ⊦ # bits = 0 ↔ bits = [];% ⊦ m < n → n ≠ 0; (forward)% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i * j = j * i;