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Proof step 21 from proof shiftright-neg-div (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 22:
2 * 0 ≤ 2 * (2 ^ (abs(j) - 2) + (i - 1) / 2), 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

fma: 2 * 0 ≤ 2 * (2 ^ (abs(j) - 2) + (i - 1) / 2)

Additional information

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