If A is any language, let ${\mathit{A}}_{\frac{\mathbf{1}}{\mathbf{3}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{3}}}$be the set of all strings in A with their middle thirds removed so that

${\mathit{A}}_{\frac{\mathbf{1}}{\mathbf{3}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{3}}}\mathbf{=}\mathbf{\left\{}\mathit{x}\mathit{z}\left|forsomey,\right|\mathit{x}|=|\mathit{y}|=|\mathit{z}\mathbf{\right|}\mathit{a}\mathit{n}\mathit{d}\mathit{x}\mathit{y}\mathit{z}\mathbf{\in }\mathit{A}\mathbf{\}}\mathbf{.}$

Show that if A is regular, then ${\mathit{A}}_{\frac{\mathbf{1}}{\mathbf{3}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{3}}}$is not necessarily regular

A language is regular if it can be expressed in terms of regular expression.A regular expression can also be described as a sequence of pattern that defines a string. Regular expressions are used to match character combinations in strings.

For the given language, $A_{\frac{1}{3}-\frac{1}{3}}=\left\{xz\left|forsomey,\right|x\left|=\right|y\left|=\right|z\right|andxyz\in A\}$Let A be any language, define $A_{\frac{1}{3}-\frac{1}{3}}$be the subset of strings of AA whose middle third is removed.

The solution is came across makes the following claim,

which cannot be justify ,

Let $A=\{0*1*\}$then $A_{\frac{1}{3}-\frac{1}{3}}$$\cap$$\{0*1*\}$$=\{0^n{1}^n\mid n⩾0\}$

For example consider the string $w=0000\#1$, removing the middle third will yield $00\#1$and whose intersection with $\left\{0^{*}{1}^{*}\right\}$is∅which not of the form$\{0^n{1}^n⩾0\}$.

A string ss belongs to$A_{\frac{1}{3}-\frac{1}{3}}$iff you can form it from a string tt of the form $0^p\#1^q$by deleting the middle third t then here the string will be in$\{0*1*\}$iff the$\#$was in the middle third of t. The question is asking you to show that if that happens, then t contains an equal number of zeros and ones.

Hence, $A_{\frac{1}{3}-\frac{1}{3}}=\left\{xz\left|forsomey,\right|x\left|=\right|y\left|=\right|z\right|andxyz\in A\}$A is regular, then $A_{\frac{1}{3}-\frac{1}{3}}$is not necessarily regular is proved.