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Proof step 2 from proof shiftright-neg-div (previous step)

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

Rule argument

name: shiftright-abs rotate: F subst: [j, i] → [j, i]

Additional information

Used terms: j, i