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Proof step 9 from proof shiftright-neg-div (previous step)

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) ⊦ i→n(abs(j) - 1) < i→n(j), 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 10:
i + 2 ^ (abs(j) - 1) < 0, i < 0, abs(j) ≤ abs(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→n(i) < i→n(j) ↔ abs(i) < abs(j);