A regular language is a language that can be represented by a regular expression(or for which there exists a DFA), then why is language represented by 0$^{+}$ not regular?

I am asking this because for the following question the correct answer is option 2

Let L denote the languages generated by the grammar S→0S0∣00. Which of the following is TRUE?

  1. L=0$^{+}$

  2. L is regular but not 0$^{+}$

  3. L is context free but not regular

  4. L is not context free

  • 2
    $\begingroup$ Because $L$ it is regular and it is not $00^*$ but $00(00)^*$, so this refers to $L$ being a different language, not to deny that $0^+$ is regular? $\endgroup$
    – Evil
    Commented Oct 12, 2017 at 14:57
  • 1
    $\begingroup$ @Evil My bad, english is not my native language. $\endgroup$
    – Sumeet
    Commented Oct 12, 2017 at 15:04
  • $\begingroup$ Please give a reference for the problem you quote. $\endgroup$
    – Raphael
    Commented Oct 12, 2017 at 16:25

1 Answer 1


The statement (2) is ambiguous. It could mean either $$ L \text{ is regular but } L \text{ is not equal to } 0^+ \tag{1} $$ Or it could mean, $$ L \text{ is regular but } 0^+ \text{ is not regular} \tag{2} $$ (1) is true and (2) is false. The authors of the question intended it to mean (1). But you thought that the meaning was (2), so you were confused.


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