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Proof step 9 from proof shiftright-pos-div (previous step)

i / 2 >> (j - 1) = (i / 2) / 2 ^ (j - 1), 0 ≤ j, 0 ≤ i, 0 < j, ∀ j, i0. i→n(i0) < i→n(i) → 0 ≤ i0 ∧ 0 ≤ j → i0 >> j = i0 / 2 ^ j ⊦ i / 2 >> (j - 1) = i / 2 ^ j

User applied rule insert equation

And obtained the following new goals

Rule argument

pos: left 1 flags: discard

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