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Proof step 15 from proof shiftright-neg-div (previous step)

¬ abs(j) - 1 < 0, 0 ≤ (i - 1) / 2 + 2 ^ (abs(j) - 1 - 1), i < 0, i + 2 ^ (abs(j) - 1) < 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) ⊦ (i - 1) / 2 >> (abs(j) - 1) = (i + 1 - 2 ^ abs(j)) / 2 ^ abs(j), j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
0 ≤ 2 ^ (abs(j) - 2) + (i - 1) / 2, i + 2 ^ (abs(j) - 1) < 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)

Rule argument

Additional information

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