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I have been at this for hours. The question is:

Prove that the language $A = \{0^kx \mid k > 0, x \in \{0,1\}^*, \text{ and } \#(0,x) \geq k\}$ is regular,

where $\#(n, x)$ denotes the number of $n$ in string $x$.

So far, I have used the pumping lemma to determine that the opposite is not regular, i.e. that if $\#(0,x) \leq k$, the language is not regular.

I have also constructed a nondeterministic finite automaton that I thought I could use to prove that this set is regular since that is generally the way to go about proving regularity. I can't really draw it, but at start state $s$, if $0$ is read in, then move to state 1 (or state $'0'$). If a $0$ is read in, either stay in state 1 or move to state 2. If a $1$ is read in, move to state 2. However, I honestly don't know what to do further.

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Hint: Show that a word belongs to $A$ iff it starts at $0$ and contains at least one more $0$. That is, $A$ is accepted by the regular expression $01^*0(0+1)^*$.

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  • $\begingroup$ Ah ok. So in essence, we are constructing a DFA with three states $s$, $1$, and $2$ where $s$ -> $1$ if a zero is read in (and an 'inaccessible' state if a one is read in), $1$ -> $1$ if a one is read in, $1$ -> $2$ if a zero is read in, and $2$ -> $2$ if either a one or a zero is read in, with $2$ being the accept state? $\endgroup$
    – tdark
    Dec 11 '14 at 6:56
  • $\begingroup$ DFAs don't have inaccessible states, but otherwise you're right. Make sure that you know why $A$ consists exactly of these words, and that you can prove it is so. $\endgroup$
    – Yuval Filmus
    Dec 11 '14 at 7:02
  • $\begingroup$ Does the proof of why A consists exactly of these words involve the DFA's delta and delta-hat functions? $\endgroup$
    – tdark
    Dec 11 '14 at 7:19
  • $\begingroup$ The proof is independent of the DFA. You have to show that a word is of the form $0^k x$, where $k > 0$ and $x$ contains at least $k$ zeroes, iff it starts with $0$ and contains at least two zeroes. $\endgroup$
    – Yuval Filmus
    Dec 11 '14 at 7:46
  • $\begingroup$ Ok. So basically I have down this: suppose for w in A that w does not begin with a zero and contain at least two 0s. Thus, we have three cases: w begins with a 1 and #(0,w) >= 2, w begins with a 1 and #(0, w) < 2, and w begins with a 0 and #(0,w) < 2. The first two cases are trivially false, as A is defined as beginning with a zero. For the third case, with w begins with a zero, k = 1 at this point in time. Since k=1, x must have, by definition, at least one zero, bringing the total number of zeroes to at least 2, thereby invalidating the case. Thus, the #(0,w) >= 2, w begins with a zero... $\endgroup$
    – tdark
    Dec 11 '14 at 8:14

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