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Proof step 20 from proof bib-pos-low (previous step)

¬(8 + 1 = 9 ∧ # int2bits(j) < 9), j ≤ 127, 64 ≤ i0 * 256 + j, lastn(8, int2bits(i0) + fillsgn(9, int2bits(j)).rest).first, 127 < i0 * 256 + j, 0 < i0, 0 ≤ j, 0 < i0 * 256 + j ⊦ 0 = i0 * 256 + j, b2i(i2b(bits2int(lastn(8, int2bits(i0) + fillsgn(9, int2bits(j)).rest)))) = j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ int2bits(j) ≠ [];% ⊦ # bits = 0 ↔ bits = [];%0 < n ⊦ (* n) ≤ m → m ≠ 0; (forward)% ⊦ n ≤ # bits → fillsgn(n, bits) = bits;% ⊦ ¬ m < n ↔ n ≤ m;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;