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Proof step 56 from proof bib-neg-high (previous step)

i0 < 0 ∧ j0 < 0 ∧ abs(j0) < 2 ^ (8 - 1) ∧ 0 < 8, i0 * 2 ^ 8 + j0 < 0, i0 ≤ 0, j0 ≤ 0, ~128 ≤ j0, i0 * 2 ^ 8 + j0 ≤ ~256, abs(j0) < 2 ^ 8 ⊦ b2i(i2b(bits2int(lastn(8, int2bits(i0 - 1) + fillsgn(i→n(8), int2bits(j0)))))) = j0, ~256 ≠ i0 * 2 ^ 8 + j0 ∧ lastn(8, int2bits(i0 - 1) + fillsgn(i→n(8), int2bits(j0))).first

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%numlslsdel-01: i0 ≠ 0, i0 < i ⊦ (* i) < j → (* i0) < j; from specification int-basic1%commutativity: ⊦ i * j = j * i; from specification int-mult%zero-gels-03: ⊦ i < 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%numls-02: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; from specification int-abs%neg-13: ⊦ ~ i < j ↔ 0 < j + i; from specification int-abs%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs