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Proof step 2 from proof shiftright-not-neg (previous step)

0 ≤ i, i >> j < 0 ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ 0 ≤ i → i >> j = i / 2 ^ abs(j)
Proof step 4:
0 ≤ i, 0 ≤ i, i / 2 ^ abs(j) < 0 ⊦

Rule argument

name: shiftright-pos-div-01 rotate: F subst: [j, i] → [j, i]

Additional information

Used terms: j, i