# Is star closure of reverse of language equivalent to reverse of closure of that language

is the following true $(L^R)^* = (L^*)^R$

I tried the following to prove it true. let u,v belong to L then $L^* = \{ u,v, uu, vv, uv, vu ... \}$ and $(L^*)^R = \{ u^R, v^R, u^Ru^R, v^Rv^R, v^Ru^R, u^Rv^R ... \}$

now $L^R = \{ u^R, v^R \}$ so $(L^R)^* = \{ u^R, v^R, u^Ru^R, v^Rv^R, u^Rv^R, v^Ru^R ... \}$

• What have you tried thus far to prove that this is true? Or have you thought about (small) counter examples? Sep 15 '12 at 8:16
• @DaveClarke : I have edited the question. I tried like this.. Sep 15 '12 at 11:02
• Kleene's star is defined inductively, so maybe an induction works (if the statement is true).
– Raphael
Sep 15 '12 at 13:01

Take any $w\in {(L^*)}^R$. Then $w$ can be writen as $w=(u_1\cdot u_2\cdots u_n)^R$, with $u_i\in L$. We have
$$w =(u_1\cdot u_2\cdots u_n)^R =u_n^R \cdot u_{n-1}^R \cdots u_1^R,$$
and therefore $w\in (L^R)^*$.
Assume now that $w\in {(L^R)}^*$, then by the same argument $$w =u_1^R \cdot u_{2}^R \cdots u_n^R= (u_n\cdot u_{n-1}\cdots u_1)^R ,$$ and hence $w\in {(L^*)}^R$.
As a consequence ${(L^*)}^R={(L^R)}^*$