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Proof step 40 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)) / (4 * 2 ^ (abs(j) - 2))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 41:
2 ^ (abs(j) - 2) = 0, i0 >> (abs(j) - 1) = (1 + i0) / 0, i0 * 2 + j1 < 0, 1 < abs(j), i0 * 2 + j1 < ~1, j1 ≤ 0, i0 ≤ 0, abs(j1) < 2, ∀ 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, (1 + i0) / 0 = (2 * (i0 + 1) + j1) / 0

Rule argument

Additional information

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