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Proof step 2 from proof shiftright-pos-div-01 (
previous step
)
0 ≤ i ⊦ i >> j = i / 2 ^ abs(j)
User applied rule insert rewrite lemma
And obtained the following new goals
Proof step 3
:
⊦ i >> j = i >> abs(j)
Proof step 4
:
0 ≤ i ⊦ i >> abs(j) = i / 2 ^ abs(j)
Rule argument
name: shiftright-abs rotate: F subst: [j, i] → [j, i]
Additional information
Used terms: j, i