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Proof step 20 from proof shiftright-neg-div (previous step)

0 < abs(j) - 1, 0 ≤ 2 ^ (abs(j) - 2) + (i - 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 ^ abs(j)) / 2 ^ abs(j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
2 * 2 ^ (abs(j) - 2) + i < 0, 1 < abs(j), 0 ≤ 2 ^ (abs(j) - 2) + (i - 1) / 2, i < 0, 2 ^ (abs(j) - 2) + i < (i - 1) / 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) ⊦ (i - 1) / 2 >> (abs(j) - 1) = (i + 1 - 2 ^ abs(j)) / 2 ^ abs(j), j = 0

Rule argument

Additional information

Used simplifier rules: ⊦ 0 < abs(i) ↔ i ≠ 0;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ 1 * i = i;% ⊦ (* i2) * i + j < i + j0 ↔ (i2 - 1) * i + j < j0;% ⊦ 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 ∧ j ≤ j0 → i < j0; (forward)% ⊦ i - i1 - j = i - i1 + j;% ⊦ i -1 = i - 1;% ⊦ i < j - i0 ↔ i + i0 < j;