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I am trying to prove that with language L, (L^R) ^R =L So that the reversal of the reversal of the language is the original language L.

I have proved that before with a string not language (let's call it a string 's').

(s^R)^R = s.
Proof by induction on the length of s.

Base: |s|=0. s=ε. (s^R)^R =(ε^R)^R =(ε)^R =ε=w.
Induction hypothesis: if 0 ≤ |x| ≤ n, then (x^R)^R = x.
Induction step: Suppose |s| = n + 1. Then we can write s = xa, where a is a symbol and 0≤|x|≤n. Then
(s^R)^R
= ((xa)^R)^R since s = xa
= (ax^R)^R since (xy)^R = y^R*x^R (think of a as a string of length 1) = (x^R)^R*a^R same as previous line but using outer R
= (x^R)^R*a using the second part of the definition of reverse with u=ε.
= xa by the induction hypothesis 
= s since s = xa

How can this reasoning the expanded to Languages? In essence I am proving the same operation, but I am dealing with languages instead of strings. Languages are made up of strings. Thoughts?

David Richerby
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user3295674
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2 Answers2

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Well, we don't really need induction here.

We want to prove that for any language $L$ holds the statement

$$\left( L^R \right)^R = L.$$

(1) Let $s \in \left( L^R \right)^R$. Then there exists a (unique) string $t \in L^R$, such that $s = t^R$. The situation is analogous with $t$: there exists a (unique) string $u \in L$ such that $t = u^R$. From that we get $s = t^R = \left( u^R \right)^R = u$, so $s \in L$.

(2) Let $s \in L$. Observe that $s^R \in L^R$ and $\left( s^R \right)^R \in \left( L^R \right)^R$, again using the fact that $s = \left( s^R \right)^R$ we get that $s \in \left( L^R \right)^R$.

From (1) and (2) follows the statement of the theorem.

Anton Trunov
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Given that it's true for strings (as you've shown), there isn't much to prove: $$\begin{align} (L^R)^R &= \{s^R\mid s\in L^R\} \\ &= \{s^R\mid s\in \{t^R\mid t\in L\}\} \\ &= \{(t^R)^R\mid t\in L\} \\ &= \{t\mid t\in L\} \\ &= L. \end{align}$$

BrianO
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