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Proof step 4 from proof shiftright-rec-pos-01 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ 0 ≤ i ∧ 0 < j → i >> j = i / 2 >> (j - 1)

Rule argument

name: shiftright-rec-pos rotate: F subst: [j, i] → [abs(j), i]

Additional information

Used simplifier rules: ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i ≤ j → (¬ ¬ j < i ↔ false);% ⊦ 0 < abs(i) ↔ i ≠ 0; Used terms: abs(j), i