Consider the language L = {0ⁱ1^j2^j3³ⁱ | i and j are ≥ 1}. You can show L is context free. It turns out Half(L) is not context free, but to prove it, it helps to intersect first with a regular language. Remember that if L were a CFL, then L intersected with any regular language would also be a CFL. Added 5/5: Actually, if you are clever about how you select z, you can use the pumping lemma directly on Half(L). You do not have to intersect with a regular language first.