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I'm working on an example which says that a string x is obtained from a string w by deleting symbols if it is possible to remove zero or more symbols from w so that just the string x remains. For example, the following strings can all be obtained from 0110 by deleting symbols:

λ, 0, 1, 00, 01, 10, 11, 010, 011, 110, and 0110.

Let Σ = {0, 1} and let A ⊆ Σ ∗ be an arbitrarily chosen regular language.

Define B = {x ∈ Σ ∗ | there exists a string w ∈ A such that x is obtained from w by deleting symbols}.

In words, a string is in B if you can obtain that string by first choosing a string from A and then deleting zero or more symbols from that chosen string. Prove that B is regular.

I'm not able to prove it. Any help would be appreciated.

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Hint 1: As $A$ is regular, it is recognized by finite automata. Try to modify this automata to make it recognize $B$.

Hint 2: Skipping "deleted" symbols might be handled by adding $\epsilon$ - transitions to the original automata in a proper way.

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Use closure properties of regular languages: Apply substitution $\sigma(x) = \{x, \epsilon\}$. This replaces symbols by themselves or nothing.

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  • $\begingroup$ should I apply this substitution on Every state of my NFA of A language? I don't understand how will it then start accepting language B. Can you please elaborate on it. Sorry, I'm new to automata theory. $\endgroup$ – jumpy123 Mar 1 at 11:48
  • $\begingroup$ @ahmad123, the substitution is applied to the language, i.e. $\sigma(L) = \{\sigma(\alpha) \colon \alpha \in L \}$. It is a theorem that regular langusgrs are closed with respect to substitutions by regular languages for each symbol, and the languagrs here are finite, thus regular. $\endgroup$ – vonbrand Mar 1 at 16:16
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$A$ has a regular expression. Can you modify the expression so it will express the optional deletion of arbitrary characters?

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