# Use closure properties to transform languages to $L := \{ a^nb^n : n\in \mathbb N \}$

For the purpose of proving that they are not regular, what closure properties can I use to transform the languages

1. $L_a = \{ a^*cw \mid w \in \{a,b \}^* \land |w|_a = |w|_b \}$ and
2. $L_b = \{ab^{i_1}ab^{i_2}\ldots ab^{i_n} \mid i_j∈\mathbb N \land \exists j∈[1,n] \ i_j \not= j \}$

to $L := \{ a^nb^n \mid n\in \mathbb N \}$, respectively?

I tried:

1. $L_a = \{ a^*cw \mid w \in \{a,b \}^* \land |w|_a = |w|_b \}$

$L_a' = \{ \{a,d\}^*cw \mid w \in \{a,b,d \}^* \land |w|_a + |w|_d = |w|_b \}$ (union?)

$L_a'' = \{ d^*cw \mid w \in \{a,b \}^* \land |w|_a = |w|_b \}$ (concatenation?)

$L_a''' = \{ w \mid w \in \{a,b \}^* \land |w|_a = |w|_b \}$ (homomorphism?)

2. $L_b = \{ab^{i_1}ab^{i_2}\ldots ab^{i_n} \mid i_j∈\mathbb N \land\exists j∈[1,n] \ i_j \not= j \}$

$L_b' = \{ab^{i_1}ab^{i_2}\ldots ab^{i_n} \mid i_j∈\mathbb N \land\forall j∈[1,n] \ i_j = j \}$ (complement?)

$L_b'' = \{ac^{i_1}ac^{i_2}\ldots ac^{i_n} \mid i_j∈\mathbb N \land\forall j∈[1,n] \ i_j = j \}$ (homomorphism?)

• What do you want to transform them into? – jmad May 20 '12 at 14:33
• Please take a little effort to formulate your question more clearly. – Dave Clarke May 20 '12 at 14:39
• I edited so that it easier to read. But I still don't get it. Do you want to prove they are not regular by saying they can be reduced to $L$ which is not? – jmad May 20 '12 at 15:10
• This looks like you are dumping your work on us without more effort than copying some assignment. Please give us some more. – Raphael May 20 '12 at 15:10
• Try to be more formal; when you write "union" write the languages which are joined, when you write "homomorphism" define the homomorphism etc. – sdcvvc May 20 '12 at 16:49

Regular languages are closed under intersection. This often allows to cut away all parts of a language that are not needed to show non-regularity. Complementation serves a similar purpose: if the original language is "complicated", the complement may be simpler so work with (in terms of other closure properties).

Hints:

For $L_a$, the part right of $c$ is sufficient; try to intersect with a regular expression that gets rid of the clutter. Note that the part right of $c$ is close to $L$. Maybe another regular expression can help?
For $L_b$, note that we get $n$ times $a$ and $b$ (in the last block), respectively, if we replace $\exists \dots i_j \neq j$ with $\forall \dots i_j=j$.

Complete solution for $L_a$:

Let $L_a' := L_a \cap \mathcal{L}(ca^*b^*) = \{ ca^nb^n \mid n \in \mathbb{N} \}$. Then, with homomorphism $\phi : \{a,b,c\} \to \{a,b\}^*$ defined by

$\qquad \displaystyle \phi(w) = \begin{cases} \varepsilon &, w = c \\ w &, \text{ else} \end{cases}$

we have $\phi(L_a') = L$.

Complete solution for $L_b$:

In order to get rid of the pesky $\exists$, we can complement. That introduces lots of words we do not want, i.e. such that don't have the $(ab^*)^*$ structure, so we can intersect with exactly that regular expression to get structure back:

$\qquad \displaystyle L_b' := \overline{L_b} \cap \mathcal{L}((ab^*)^*) = \{ abab^2ab^3\dots ab^n \mid n \in \mathbb{N} \}$.

Note that words in $L_b'$ contain exactly $n$ times the symbol $a$; if we can get rid of all blocks of $b$ but the last, we have $L$. This is possible by (nondeterministic) finite state transduction which $\mathsf{REG}$ is closed under. The transducer removes all $b$ until it decides on an $a$ that is was the last one. After that it emits the input and accepts if no further $a$ is encountered.

Note that if we reverse $L_b'$, we can cut away the excess $b$ with a deterministic transducer and reverse again to obtain $L$.

Hint: for $L_a$ you can use stability with intersection from $L_a'''$.

1. $L_a = \{ a^*cw \mid w \in \{a,b \}^* \land |w|_a = |w|_b \}$

$L_a \cup d^* = L_a'$

$L_a' = \{ \{a,d\}^*cw \mid w \in \{a,b,d \}^* \land |w|_a + |w|_d = |w|_b \}$ (union?)

$L_a' \cap a^*cw = L_a''$

$L_a'' = \{ d^*cw \mid w \in \{a,b \}^* \land |w|_a = |w|_b \}$

homomorphism $h(d) = \epsilon, h(c) = \epsilon, h(a) = a, h(b) = b$

$L_a''' = \{ w \mid w \in \{a,b \}^* \land |w|_a = |w|_b \}$

$L_a''' = L := \{ a^nb^n \mid n\in \mathbb N \}$ is the same but other syntax?

2. $L_b = \{ab^{i_1}ab^{i_2}\ldots ab^{i_n} \mid i_j∈\mathbb N \land\exists j∈[1,n] \ i_j \not= j \}$

$\overline L_b = L_b'$ so follows $\forall j∈[1,n] \ i_j = j$

$L_b' = \{ab^{i_1}ab^{i_2}\ldots ab^{i_n} \mid i_j∈\mathbb N \land\forall j∈[1,n] \ i_j = j \}$ (complement?)

homomorphism $h(b) = c, h(a) = a$ i say b is know my new c - is my idea of thislanguage correct? - $ab abb abbb ... ab^n$ ? O.o

$L_b'' = \{ac^{i_1}ac^{i_2}\ldots ac^{i_n} \mid i_j∈\mathbb N \land\forall j∈[1,n] \ i_j = j \}$

$L_b'' \cap a^*b^*$ and than homorphism $h(a) = a, h(b)=b, h(c) = \epsilon \\\$ first we have $a \epsilon ^{i_1}a \epsilon ^{i_2}\ldots a \epsilon ^{i_n}$ and than $a^*b^*$. Can anyone give me a hint for the transformation to $a^nb^n$?

• – Raphael May 23 '12 at 20:42
• Note that $L_b'$ is not the complement of $L_b$! For instance, $b \notin L_b$ but $b \notin L_b'$, either. – Raphael May 24 '12 at 21:51