# Question & Answer: For a word x, we use x^r to denote its reverse (e.g., the reverse of abaac is caaba). For a…..

For a word x, we use x^r to denote its reverse (e.g., the reverse of abaac is caaba). For a language L, we use L^r to denote {x^r: x elementof L}. Show that if L is regular then so is L^r.

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Hi,

Given

reverse for a word x= x^r

Reverse of a language L = L^r where {x^r belongs to L}

It is one of the topic in finite automata, from above statements we can say that if L can be written as a regular expression , then L^r can also be written as a regular expression.

Regular expression is built from smaller one by union , concatenation and Kleen star.

Also we can consider the closure properties of regular languages unse simple set operations .

First through Union :

$\left ( L1\cup L2 \right )^R= L1^R\cup L2^R; (L1L2)^R=L2^RL1^R; (L1^*)^R=(L1^R)^*$.

Say a word x is a1a2a3…an , reversal of it is written backwards $a_{n}a_{n-1}.......a_{1}.$

$E=F^*. Then E^R=(F^R)$

We use x^R for reversal of word x $X\epsilon L(E)$, each $X_{n}^R$ is in L(F). The reversal of a string in L(F^*) . Which is in L(E).