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Letxandy be strings and let L be any language. We say that x and y are distinguishable by L if some string Z exists whereby exactly one of the stringsxzandyz is a member of L ; otherwise, for every string z , we have xzLwhenever yzLand we say that are indistinguishable by L. If xandyare indistinguishable by L, we write x 鈮 y. Show thatLis an equivalence relation.

Short Answer

Expert verified

The relation L is reflexive, symmetric and transitive.

Step by step solution

01

Define Equivalence relation

An equivalence relation is a relationship on a set, generally denoted by 鈥溾埣鈥, that is reflexive, symmetric, and transitive for everything in the set.

The relation 鈥渋s equal to鈥, denoted= , is an equivalence relation on the set of real numbers since for any

1. (Reflexivity) x=x

2. (Symmetry) if x=ytheny=x

3. (Transitivity) if x=yandy=zthenx=z

02

Prove that the relation is reflexive

Consider an alphabet.Letx*;Fory

xyLifanonlyifxyL

So, X is indistinguishable to X and X is arbitrary.

Therefore, Lis reflexive

03

Prove that the relation is symmetric

Suppose xLy . This implies that for all z*,

  1. xz,yzLorxz,yzLforallz*.
  2. yz,xzLoryz,xzLforallz*

The implication in 2meansyLx.

Thus, Lis symmetric.

04

Prove that the relation is transitive

Consider x,y,z*,xLy,andyLz.

To prove transitivityshow that xLzoru*,xuLif and onlyifzuL. Consider that u*andxuL. Now, yuLandzuLbecausexLyandyLzyrespectively. Also, xuL.FromxLy,yLz,it is concluded that yuLandzuL.xuLif and only if zuL.ItistruethatxLz.

Because x,y,andzare arbitrary, Lis transitive.

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