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Proof step 1 from proof shiftright-rec-pos-01 (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)