Regular/Non-regular Language

Question

Given a set $A\subseteq \left\{0,1 \right \}^*,$ let

$A' = \left\{ xy\; |\;x1y\in A\right\}.$ That is, $A'$ consistes of all the strings obtained from a string in $A$ by deleting in $A$ by

deleting exactly one $1$. if $A$ is regular, then $A'$ is 

(A) Regular

(B) Context free but not regular

(C) Recursive but not context free

(D) None of the above.

Answer 1

$A$ is regular 

Design a $DFA$ $M1$ for language $A$. 

Design a $DFA$ $M2$ for $(0+1)^*1(0+1)^*$

Do the Intersection, using $M1\times M2$  

Now,We will have $DFA$  $M$ that accepts all strings $x1y\in A$

There will be at least one transition in $M$ such as $q \times 1 \rightarrow q'$,where $q \neq q'$ ,while moving from start state to final state, and not in closure (not is closed loop, $x1y$ guaranteed for such transition), replace $1$ by $\epsilon$ in that one transition as $q \times \epsilon \rightarrow q'$  
so , after converting $\epsilon -NFA$ to $DFA$, that will be  $DFA$ of $A'$

$A'$ will be Regular

Comments

Nice :)

Answer 2

A is regular

A' is deletes only one 1 (deleting finite no. of string)

A'=A - 1

   =regular - finite amount of string

  =regular -regular

  =regular ⋂ (regular)'

  =regular ⋂ regular

  =regular

Ans will be (A)

Comments

1 is deleted from every string in A. It is not that we are removing strings containing 1 from A - even then we might be removing infinite number of strings.

So, your logic though good is not correct for this question.