Question

Let it$B$ be any language over the alphabet $\sum$. Prove that $B=B+iffBB\subseteq B$

Short answer

Thus, the solution is$x_1,x_2,x_3\in B$ .

Step-by-step solution

Introduction

‘B’ is apply to any language over that alphabet$\text{∑}$.

To be confirmation about formula :-

$B=B^{+}$iff $BB\subseteq B$

Explanation

One Direction: -

Assume: - ^{$B=B^{+}$} ----------- result (1)

To Show : -$BB\subseteq B$

Since, forevery language $BB\subseteq B^{+}$ ------- result (2)

By Sabstitating (1) and (2) , it can be obtained that $BB\subseteq B$. Hence, it’s been confirmed that $BB\subseteq B$ iff role="math" localid="1663234619808" $BB=B^{+}$

Other Direction : -

Assume : - $BB\subseteq B$

To sure and confirmed : - $-BB=B^{+}$

It’s in knowledge in every or any language $BB\subseteq B^{+}$ --------- (3)

Let w be a string of elements, such that $w\in B^{+}w\in B$.

If $w\in B+$than W the string can be divided within elements. $x_1,x_2,x_3,.....x_{k}suchasw=x_1,x_2,x_3....x_k$For $x_1\in Bandk⩾1$ .

After long duration $x_1{x}_2\in B$and it is assumed that $BB\in B.$

Same way, it hold for the different elements $x_3$, as well as $x_3\in B$and $BB\subseteq B$.Thus

$x_1,x_2,x_3\in B$ .