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Proof step 21 from proof bib-neg-high (previous step)

i0 < 0, i0 * 256 + j0 < 0, i0 ≤ 0, j0 ≤ 0, ~128 ≤ j0, i0 * 256 + j0 ≤ ~256, abs(j0) < 256 ⊦ b2i(i2b(bits2int(lastn(8, int2bits(i0 * 256 + j0))))) = j0, ~256 ≠ i0 * 256 + j0 ∧ lastn(8, int2bits(i0 * 256 + j0)).first, i0 < 0 ∧ j0 < 0 ∧ ~128 ≠ j0

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%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%zero-gels-03: ⊦ i < 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%commutativity: ⊦ i * j = j * i; from specification int-mult