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Proof step 37 from proof shiftright-neg-div (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 38:
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)) / (4 * 2 ^ (abs(j) - 2))

Rule argument

Additional information

Used simplifier rules: ⊦ i < (* j) ∧ j < 0 → i < 0; (forward)% ⊦ i * j + i * j0 = i * (j + j0);% ⊦ i + (* j) < (* j0) ↔ i < j0 - j;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j + i = i * (j + 1);% ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward)% ⊦ i < 0 ∧ 0 ≤ j → i ≤ j;% ⊦ i < 1 ↔ i ≤ 0;% ⊦ i - j < i0 ↔ i < j + i0;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i - j = i0 ↔ i = i0 + j;