TOC PYQ 2019 Solution

Question 1 (a): Prove that for all sets S, (S+)=S.

To prove that for all sets S, (S+)=S, we need to show that (S+)* is a subset of S* and S* is a subset of (S+)*.

Let’s start by proving that (S+)* is a subset of S*.

(S+)* is the set of all strings that can be formed by concatenating one or more strings from S. S* is the set of all strings that can be formed by concatenating zero or more strings from S. Since (S+)* includes all the strings in S*, (S+)* is a subset of S*.

Now let’s prove that S* is a subset of (S+)*.

S* is the set of all strings that can be formed by concatenating zero or more strings from S. Since the empty string is included in S*, it is also included in (S+). Therefore, S is a subset of (S+)*.

Since (S+)* is a subset of S* and S* is a subset of (S+), we can conclude that (S+)=S* for all sets S.

Question 1 (b): Give regular expression for the language of all strings that do not end with double letter.

The regular expression for the language of all strings that do not end with a double letter is ((a|b)*|(a|b)*ab|(a|b)*ba) .

This regular expression can be broken down into three parts:

• (a|b)*: This part matches all strings that end with a single letter.
• (a|b)*ab: This part matches all strings that end with an a followed by a b.
• (a|b)*ba: This part matches all strings that end with a b followed by an a.

By combining these three parts using the | operator, we can match all strings that do not end with a double letter.

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