Is Context Free Language closed under perfect shuffle?

Note that this is not shuffle but perfect shuffle, defined as follows:

Let $w = a_{1}a_{2} \ldots a_{n}$ and $x = b_{1}b_{2} \ldots b_{n}$ be two strings of the same length. Then the perfect shuffle of $w$ and $x$ is defined as $a_{1}b_{1}a_{2}b_{2} \ldots a_{n}b_{n}$.

So the question is: are context free languages closed under perfect shuffle?

• Do you define the perfect shuffle as an unary or binary operation? I.e. are $w$ and $x$ from the same regular language or from two different regular languages? Nov 2 '14 at 23:48
• So this is not actually shuffling but interleaving. Nov 3 '14 at 0:52
• No. Actually a in w and b in x are both characters. So w and x are both strings formed by ai and bi. Nov 3 '14 at 1:26
• @DavidRicherby Interleaving seems good terminology, but perfect shuffle is sometimes used. Nov 3 '14 at 22:00

Nope. Shuffle $\{ a^n b^{2n} \mid n\ge1\}$ with $\{ a^{2n} b^n \mid n\ge1\}$ .
• Its almost the same proof as for $\{a^n b^n c^n \mid n \geq 1\}$ Nov 3 '14 at 7:29