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Proof step 56 from proof shiftright-neg-div (previous step)

i0 >> (abs(j) - 1) = (i0 + 1 - 2 * 2 ^ (abs(j) - 2)) / (2 * 2 ^ (abs(j) - 2)), 2 * 2 ^ (abs(j) - 2) ≤ i0 + 1, 1 < abs(j), i0 + 2 ^ (abs(j) - 2) < 0, i0 < 0, ∀ i, j0. i→n(j0) < i→n(j) → i + 2 ^ (abs(j0) - 1) < 0 → j0 = 0 ∨ i >> abs(j0) = (i + 1 - 2 ^ abs(j0)) / 2 ^ abs(j0) ⊦ j = 0, abs(j) = 1, 2 ^ (abs(j) - 2) = 0, i0 >> (abs(j) - 1) = ((2 * (i0 + 1 - 2 * 2 ^ (abs(j) - 2)) - 1) / 2) / (2 * 2 ^ (abs(j) - 2))

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