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Proof step 7 from proof shiftright-neg-div (previous step)

i < 0 ∧ abs(j) ≠ 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) ⊦ j = 0 ∨ (i - 1) / 2 >> (abs(abs(j)) - 1) = (i + 1 - 2 ^ abs(j)) / 2 ^ abs(j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
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: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ abs(i) = 0 ↔ i = 0;% ⊦ abs(abs(i)) = abs(i);