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This exercise come from Sipser (1.68).

In essence, show that if $A$ is regular, then $CUT(A) := \{yxz | xyz \in A\}$ is regular. I've managed to show that $B = \{yx | xyz \in A\}$ is regular. Furthermore, I've shown that $SUFFIX(A) = \{z | xz \in A\}$ is regular. My idea was to then say that $CUT(A)$ is precisely the concatenation of $B$ and $SUFFIX(A)$. However, the trouble I'm running into is that this operation will attach the "wrong" suffixes to a string. For instance, if $A$ is "strings in alphabetical order", then $bad$ will be in the concatenation of $B$ and $SUFFIX(A)$.

How can I complete this?

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  • $\begingroup$ If you are not familiar with the guessing technique, please read the parts of this answer of Yuval on how to prove a language is regular that explains that technique. $\endgroup$
    – John L.
    Jan 27, 2022 at 18:41
  • $\begingroup$ How did you show that $\{yx \mid xyz \in A\}$ is regular? If you used a construction with automata it can probably be extended to include the suffix $z$? $\endgroup$ Jan 28, 2022 at 2:24

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