Understanding Regular Expressions and Their Languages

This problem asks us to compare the languages generated by three different regular expressions: $r = a(a + b)^*$, $s = aa^*b$, and $t = a^*b$. We need to determine the truthfulness of two statements regarding the subset relationships between these languages, denoted as $L(r)$, $L(s)$, and $L(t)$.

Defining the Languages

Let's first understand what each regular expression represents:

• Regular Expression $r = a(a + b)^*$: This expression matches strings that start with the character 'a' followed by any sequence of 'a's and 'b's, including an empty sequence. The language $L(r)$ includes strings like: 'a', 'aa', 'ab', 'aaa', 'aab', 'aba', 'abb', etc. Essentially, any string starting with 'a'.
• Regular Expression $s = aa^*b$: This expression matches strings that start with 'a', followed by zero or more 'a's ($a^*$), and finally ending with 'b'. The language $L(s)$ includes strings like: 'ab' (when $a^*$ is empty), 'aab' (when $a^*$ is one 'a'), 'aaab' (when $a^*$ is two 'a's), and so on. These strings are of the form $a^n b$ where $n \ge 1$.
• Regular Expression $t = a^*b$: This expression matches strings that start with zero or more 'a's ($a^*$) followed by 'b'. The language $L(t)$ includes strings like: 'b' (when $a^*$ is empty), 'ab' (when $a^*$ is one 'a'), 'aab' (when $a^*$ is two 'a's), etc. These strings are of the form $a^m b$ where $m \ge 0$.

Analyzing Statement (i): $L(s) \subseteq L(r)$ and $L(s) \subseteq L(t)$

Statement (i) has two parts. Let's examine each one:

1. Is $L(s) \subseteq L(r)$?
   This asks if every string in $L(s)$ is also in $L(r)$.
   • Strings in $L(s)$ are of the form $a^n b$ where $n \ge 1$. Examples: 'ab', 'aab', 'aaab'.
   • Strings in $L(r)$ start with 'a' and are followed by any combination of 'a' and 'b'. Examples: 'a', 'aa', 'ab', 'aab', 'aba', 'abb'.
   • Consider a string in $L(s)$, say $w = a^n b$ where $n \ge 1$. Since $n \ge 1$, the string $w$ must start with at least one 'a'. After the initial 'a', the rest of the string ($a^{n-1}b$) is a sequence of 'a's and 'b's. Thus, $w$ fits the definition of a string in $L(r)$ (starts with 'a' followed by any sequence).
   • Therefore, every string in $L(s)$ is also in $L(r)$. $L(s) \subseteq L(r)$ is True.
2. Is $L(s) \subseteq L(t)$?
   This asks if every string in $L(s)$ is also in $L(t)$.
   • Strings in $L(s)$ are of the form $a^n b$ where $n \ge 1$. Examples: 'ab', 'aab', 'aaab'.
   • Strings in $L(t)$ are of the form $a^m b$ where $m \ge 0$. Examples: 'b', 'ab', 'aab', 'aaab'.
   • Consider a string in $L(s)$, say $w = a^n b$ where $n \ge 1$. This string is of the form $a$s followed by a 'b', with at least one 'a'. This matches the form $a^m b$ where $m \ge 0$, specifically with $m=n$.
   • Therefore, every string in $L(s)$ is also in $L(t)$. $L(s) \subseteq L(t)$ is True.

Since both parts of statement (i) are true, statement (i) is Correct.

Analyzing Statement (ii): $L(r) \subseteq L(s)$ and $L(s) \subseteq L(t)$

Statement (ii) also has two parts. Let's examine each one:

1. Is $L(r) \subseteq L(s)$?
   This asks if every string in $L(r)$ is also in $L(s)$.
   • Strings in $L(r)$ start with 'a' and can be followed by any combination of 'a' and 'b'. Examples: 'a', 'aa', 'ab', 'aaa', 'aab', 'aba', 'abb'.
   • Strings in $L(s)$ are of the form $a^n b$ where $n \ge 1$. Examples: 'ab', 'aab', 'aaab'. These strings must end with 'b'.
   • Consider the string 'a'. The string 'a' is in $L(r)$. Is 'a' in $L(s)$? No, because strings in $L(s)$ must end with 'b'.
   • Consider the string 'aa'. The string 'aa' is in $L(r)$. Is 'aa' in $L(s)$? No, because strings in $L(s)$ must end with 'b'.
   • Consider the string 'aba'. The string 'aba' is in $L(r)$. Is 'aba' in $L(s)$? No, because strings in $L(s)$ are only sequences of 'a's followed by a single 'b' (like $a^n b$).
   • Since there are strings in $L(r)$ that are not in $L(s)$, $L(r) \subseteq L(s)$ is False.
2. Is $L(s) \subseteq L(t)$?
   This is the same check as the second part of statement (i). As determined before, this is True.

Since the first part of statement (ii) is false, statement (ii) is Incorrect.

Summary of Findings

Based on the analysis of the languages $L(r)$, $L(s)$, and $L(t)$ derived from the regular expressions $r = a(a + b)^*$, $s = aa^*b$, and $t = a^*b$, we found:

• Statement (i) ($L(s) \subseteq L(r)$ and $L(s) \subseteq L(t)$) is Correct.
• Statement (ii) ($L(r) \subseteq L(s)$ and $L(s) \subseteq L(t)$) is Incorrect.

Therefore, only statement (i) is correct.

Conclusion on Regular Expression Language Comparison

Comparing the languages $L(r)$, $L(s)$, and $L(t)$ reveals specific subset relationships. We found that $L(s)$ is a subset of both $L(r)$ and $L(t)$, making statement (i) true. However, $L(r)$ is not a subset of $L(s)$, making statement (ii) false. Thus, only statement (i) holds true.

Revision Table: Regular Expression Language Properties

Additional Information: Exploring Regular Expression Languages

Regular expressions are a powerful tool in computer science for pattern matching and describing formal languages, specifically regular languages. Understanding the language $L(\text{RegEx})$ is crucial when working with regular expressions.

The operators used in regular expressions—union ($+$), concatenation (juxtaposition), and Kleene star ($*$)—correspond directly to operations on the languages they represent:

• The language of the union of two regular expressions $R_1$ and $R_2$ is the union of their individual languages: $L(R_1 + R_2) = L(R_1) \cup L(R_2)$.
• The language of the concatenation of two regular expressions $R_1 R_2$ is the concatenation of their individual languages: $L(R_1 R_2) = L(R_1)L(R_2) = \{uv \mid u \in L(R_1), v \in L(R_2)\}$.
• The language of the Kleene star of a regular expression $R^*$ is the Kle Kleene star of its language: $L(R^*) = (L(R))^*$, which is the set of all strings formed by concatenating zero or more strings from $L(R)$. This includes the empty string $\epsilon$.

In our problem:

• $L(r) = L(a(a+b)^*) = L(a)L((a+b)^*) = \{a\} (\{a\} \cup \{b\})^*$. This is the set of all strings starting with 'a'.
• $L(s) = L(aa^*b) = L(a)L(a^*)L(b) = \{a\}\{a\}^*\{b\} = \{a^n b \mid n \ge 1\}$. (Note: $L(a^*) = \{a^m \mid m \ge 0\}$, so $L(a)L(a^*) = \{a\} \{a^m \mid m \ge 0\} = \{a^{m+1} \mid m \ge 0\} = \{a^k \mid k \ge 1\}$. Then $\{a^k \mid k \ge 1\} \{b\} = \{a^k b \mid k \ge 1\}$).
• $L(t) = L(a^*b) = L(a^*)L(b) = \{a\}^*\{b\} = \{a^m \mid m \ge 0\}\{b\} = \{a^m b \mid m \ge 0\}$.

Comparing these formal definitions confirms our string-based analysis of the subset relationships between $L(r)$, $L(s)$, and $L(t)$.