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Proof step 35 from proof shiftright-neg-div (previous step)

0 < abs(j) - 1, (i - 1) / 2 >> (abs(j) - 1) = ((i - 1) / 2 + 1 - 2 ^ (abs(j) - 1)-1 * 2) / (2 ^ (abs(j) - 1)-1 * 2), i + 2 ^ (abs(j) - 1)-1 * 2 < 0, 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) ⊦ j = 0, (i - 1) / 2 >> (abs(j) - 1) = (i + 1 - 2 * 2 ^ (abs(j) - 1)-1 * 2) / (2 * 2 ^ (abs(j) - 1)-1 * 2), abs(j) = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 36:
(i - 1) / 2 >> (abs(j) - 1) = ((i - 1) / 2 + 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) ⊦ (i - 1) / 2 >> (abs(j) - 1) = (i + 1 - 4 * 2 ^ (abs(j) - 2)) / (4 * 2 ^ (abs(j) - 2)), j = 0, abs(j) = 1

Rule argument

Additional information

Used simplifier rules: ⊦ 0 < abs(i) ↔ i ≠ 0;% ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward)% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ i < j - i0 ↔ i + i0 < j;% ⊦ i - i1 - j = i - i1 + j;% ⊦ i -1 = i - 1;