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Proof step 55 from proof shiftright-neg-div (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 56:
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))

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i ≤ j;% ⊦ 0 < i → (i * j < 0 ↔ j < 0);% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i < 1 ↔ i ≤ 0;% ⊦ 0 < i ∧ 0 ≤ j → ((* i) * i0 < (* j) ↔ i0 < (i + j - 1) / i);% ⊦ i - j < i0 ↔ i < j + i0;%i < 0 ⊦ (* i) + j = j - ~ i;% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ i < 0 → (~2 < i ↔ i = ~1);% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0;% ⊦ ~ i < j ↔ 0 < j + i;% ⊦ i < 0 → abs(i) = ~ i;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i ≠ j → (i ≤ j ↔ i < j);