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Proof step 59 from proof shiftright-neg-div (previous step)

¬(i0 + 1 < 0 ∧ 0 < 2 * 2 ^ (abs(j) - 2)), 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 + 1 - 2 * 2 ^ (abs(j) - 2)) / (2 * 2 ^ (abs(j) - 2)) = ((2 * (i0 + 1 - 2 * 2 ^ (abs(j) - 2)) - 1) / 2) / (2 * 2 ^ (abs(j) - 2))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 60:
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)), i0 ≠ ~1 ∧ ¬ abs(j) < 2

Rule argument

Additional information

Used simplifier rules: ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j;% ⊦ 0 < 2 ^ i ↔ 0 ≤ i;% ⊦ 0 < i → (0 < i * j ↔ 0 < j);% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i < 0 → (~1 ≤ i ↔ i = ~1);% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i + (* j) < (* j0) ↔ i < j0 - j;