\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
[tex]$p$[/tex] & [tex]$q$[/tex] & [tex]$r$[/tex] & [tex]$p \leftrightarrow q$[/tex] & [tex]$q \leftrightarrow r$[/tex] & [tex]$p \wedge q$[/tex] & [tex]$(p \leftrightarrow q) \wedge (q \leftrightarrow r) \rightarrow p \wedge q$[/tex] \\
\hline
T & T & T & & & & \\
\hline
T & T & F & & & & \\
\hline
T & F & T & & & & \\
\hline
T & F & F & & & & \\
\hline
F & T & T & & & & \\
\hline
F & T & F & & & & \\
\hline
F & F & T & & & & \\
\hline
F & F & F & & & & \\
\hline
\end{tabular}



Answer :

Certainly! Let's fill in the truth table step by step to arrive at the final column. We will evaluate each expression for the given combinations of [tex]\( p \)[/tex], [tex]\( q \)[/tex], and [tex]\( r \)[/tex].

#### Step-by-Step Truth Table Analysis:
1. Columns [tex]\( p \leftrightarrow q \)[/tex] and [tex]\( q \leftrightarrow r \)[/tex]:
- The biconditional [tex]\( p \leftrightarrow q \)[/tex] is true if and only if [tex]\( p \)[/tex] and [tex]\( q \)[/tex] have the same truth value.
- Similarly, [tex]\( q \leftrightarrow r \)[/tex] is true if and only if [tex]\( q \)[/tex] and [tex]\( r \)[/tex] have the same truth value.

2. Column [tex]\( p \wedge q \)[/tex]:
- The conjunction [tex]\( p \wedge q \)[/tex] is true if and only if both [tex]\( p \)[/tex] and [tex]\( q \)[/tex] are true.

3. Column [tex]\((p \leftrightarrow q) \wedge (q \leftrightarrow r)\)[/tex]:
- This column is the conjunction of the evaluations of [tex]\( p \leftrightarrow q \)[/tex] and [tex]\( q \leftrightarrow r \)[/tex].

4. Final Column [tex]\((p \leftrightarrow q) \wedge (q \leftrightarrow r) \rightarrow p \wedge q \)[/tex]:
- This column is true if the implication [tex]\((p \leftrightarrow q) \wedge (q \leftrightarrow r) \rightarrow p \wedge q\)[/tex] is true. An implication [tex]\( A \rightarrow B \)[/tex] is true whenever [tex]\( A \)[/tex] is false or [tex]\( B \)[/tex] is true.

Now, let's fill the table using these logical rules.

#### Constructed Truth Table:
[tex]\[ \begin{array}{|c|c|c|c|c|c|c|} \hline p & q & r & p \leftrightarrow q & q \leftrightarrow r & p \wedge q & (p \leftrightarrow q) \wedge (q \leftrightarrow r) \rightarrow p \wedge q \\ \hline \text{T} & \text{T} & \text{T} & \text{T} & \text{T} & \text{T} & \text{T} \\ \text{T} & \text{T} & \text{F} & \text{T} & \text{F} & \text{T} & \text{T} \\ \text{T} & \text{F} & \text{T} & \text{F} & \text{F} & \text{F} & \text{T} \\ \text{T} & \text{F} & \text{F} & \text{F} & \text{T} & \text{F} & \text{T} \\ \text{F} & \text{T} & \text{T} & \text{F} & \text{T} & \text{F} & \text{T} \\ \text{F} & \text{T} & \text{F} & \text{F} & \text{F} & \text{F} & \text{T} \\ \text{F} & \text{F} & \text{T} & \text{T} & \text{F} & \text{F} & \text{T} \\ \text{F} & \text{F} & \text{F} & \text{T} & \text{T} & \text{F} & \text{T} \\ \hline \end{array} \][/tex]

This filled truth table gives us a complete analysis of the logical expression [tex]\((p \leftrightarrow q) \wedge (q \leftrightarrow r) \rightarrow p \wedge q\)[/tex] for all possible truth values of [tex]\( p \)[/tex], [tex]\( q \)[/tex], and [tex]\( r \)[/tex].