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Proof step 43 from proof shiftright-neg-div (previous step)

(4 * 2 ^ (abs(j) - 2)) % 2 = 0 ∧ 4 * 2 ^ (abs(j) - 2) ≠ 0, i0 >> (abs(j) - 1) = (i0 + 1 - 2 * 2 ^ (abs(j) - 2)) / (2 * 2 ^ (abs(j) - 2)), abs(j1) < 2, i0 ≤ 0, j1 ≤ 0, i0 * 2 + j1 < ~1, 1 < abs(j), 2 * (i0 + 2 ^ (abs(j) - 2)) + j1 < ~1, i0 * 2 + j1 < 0, 2 * (i0 + 2 ^ (abs(j) - 2)) + j1 < 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) ⊦ abs(j) = 1, j = 0, i0 >> (abs(j) - 1) = ((2 * (i0 + 1) + j1 - 4 * 2 ^ (abs(j) - 2)) / 2) / (4 * 2 ^ (abs(j) - 2)) / 2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 44:
i0 >> (abs(j) - 1) = (i0 + 1 - 2 * 2 ^ (abs(j) - 2)) / (2 * 2 ^ (abs(j) - 2)), abs(j1) < 2, i0 ≤ 0, j1 ≤ 0, i0 * 2 + j1 < ~1, 1 < abs(j), 2 * (i0 + 2 ^ (abs(j) - 2)) + j1 < ~1, i0 * 2 + j1 < 0, 2 * (i0 + 2 ^ (abs(j) - 2)) + j1 < 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) ⊦ abs(j) = 1, j = 0, i0 >> (abs(j) - 1) = ((2 * (i0 + 1) + j1 - 4 * 2 ^ (abs(j) - 2)) / 2) / (4 * 2 ^ (abs(j) - 2)) / 2, 2 ^ (abs(j) - 2) = 0

Rule argument

Additional information

Used simplifier rules: ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i * j = 0 ↔ i = 0 ∨ j = 0;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;%i0 % j = 0, j ≠ 0 ⊦ (i * (* i0)) % (* j) = 0;