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Proof step 18 from proof shiftright-neg-div (previous step)

¬ 0 < abs(j) - 1, 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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
0 ≤ 2 ^ (abs(j) - 2) + (i - 1) / 2, i + 2 ^ (abs(j) - 1) < 0, i < 0, abs(j) ≤ 1, ∀ 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 < j ↔ j ≤ i;% ⊦ i < j - i0 ↔ i + i0 < j;