Is language $\{a,b\}^*$ same as language $\{xy \in \{a,b\}^* \mid |x|_a = |y|_b \}$?

Question

I need to prove or disprove if these two languages are same. So I assume that these lanaguages are same because I think that every word from $\{a,b\}^*$ could be concatenated from two words $x$ and $y$ where $|x|_a = |y|_b$. So $x$ and $y$ must belong to $\{a,b\}^*$, too. Then I can write and prove this:

$\{a,b\}^* \subseteq \{xy \in \{a,b\}^* \mid |x|_a = |y|_b \} \wedge \{xy \in \{a,b\}^* \mid |x|_a = |y|_b \} \subseteq \{a,b\}^*$

So I assume that $z \in \{a,b\}^*$. Then $z = z_1z_2...z_i$ where $z_1,z_2,...,z_i \in \{a,b\}$

Here I stucked and I don't know (1) how to continue, (2) if the way of proving i choose is correct or not.

Answer 1

You can prove this by induction on the length of the string.

The base case of $|z| = 0$ is obvious, we have to take $x=y=\varepsilon$.

The inductive assumption then is that $L =\{xy \in \{a,b\}^{\ast}\mid |x|_{a} = |y|_{b}\}$ contains all strings¹ of length at most $n$.

Now we want to show that it contains all strings of length $n+1$.

Let $z \in \{a,b\}^{n+1}$ and let $z^{-1}$ be $z$ with the final symbol removed. By the inductive assumption $z^{-1} \in L$. The final symbol $z_{n+1}$ in $z$ must be either an $a$ or a $b$.

1. If $z_{n+1}$ is an $a$, then we can construct an appropriate $x_{z}$ and $y_{z}$ by taking $x_{z} = x_{z^{-1}}$ and $y_{z} = y_{z^{-1}}a$. Then $|x_{z}|_{a} = |x_{z^{-1}}|_{a} = |y_{z^{-1}}|_{b} = |y_{z}|_{b}$, and therefore $z\in L$.

2. If $z_{n+1}$ is a $b$ we have a slightly more complicated situation, but it's not too difficult. The first symbol of $y_{z^{-1}}$ must also be either an $a$ or a $b$, or $y = \varepsilon$.

   • If the first symbol is an $a$, then we use the same $x$-$y$ partition as for $z^{-1}$, but we move the first symbol of $y$ to the end of $x$, and add the new $b$ to the end of $y$, then $|x_{z}|_{a} = |x_{z^{-1}}|_{a} + 1$ and $|y_{z}|_{b} = |y_{z^{-1}}|_{b} + 1$.

   • If the first symbol is a $b$, we do the same as in the $a$ case, but as we're adding a $b$ to $x$ and taking one $b$ away from $y$ but adding another at the end, the number of $a$s and $b$s in $x$ and $y$ don't change.

   • If $y_{z^{-1}} = \varepsilon$, then we simply choose $x = z$ and $y = \varepsilon$ and again $|x_{z}|_{a} = |x_{z^{-1}}|_{a} = |y_{z^{-1}}|_{b} = |y_{z}|_{b}$ (in fact, they all equal $0$, so $z = b^{n+1}$).

Therefore in all cases $z \in L$, and we are done.

Footnotes

1. ...over alphabet $\Sigma = \{a,b\}$.

Answer 2

We want to show that every string $z\in \{a,b\}^*$ is of the form $xy$ with $|x|_a=|y|_b$, as follows.

Make a graph of the value $|x|_a - |y|_b$ for every composition $x,y$ of $z = xy$, letting $x$ run over the prefixes of $z$. The value starts at $-|z|_b$ for $x=\varepsilon$ and ends in $|z|_a$ for $x=z$. Carefully argue it will be zero somewhere on the way.