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Proof step 1 from proof shiftright-neg-div (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
i + 2 ^ (abs(j) - 1) < 0 ⊦ j = 0, i >> j = (i + 1 - 2 ^ abs(j)) / 2 ^ abs(j)

Rule argument

Additional information

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