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.
Expert Answer
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 :
.
Say a word x is a1a2a3…an , reversal of it is written backwards
We use x^R for reversal of word x , each
is in L(F). The reversal of a string in L(F^*) . Which is in L(E).