Biconditional is Transitive/Formulation 1/Proof 2

Theorem

 * $p \iff q, q \iff r \vdash p \iff r$

Proof
We apply the Method of Truth Tables.

As can be seen for all models by inspection, where the truth values under the main connective on the LHS is $T$, that under the one on the RHS is also $T$:

$\begin{array}{|ccccccc||ccc|} \hline (p & \iff & q) & \land & (q & \iff & r) & p & \iff & r \\ \hline F & T & F & T & F & T & F & F & T & F \\ F & T & F & F & F & F & T & F & F & T \\ F & F & T & F & T & F & F & F & T & F \\ F & F & T & F & T & T & T & F & F & T \\ T & F & F & F & F & T & F & T & F & F \\ T & F & F & F & F & F & T & T & T & T \\ T & T & T & F & T & F & F & T & F & F \\ T & T & T & T & T & T & T & T & T & T \\ \hline \end{array}$

Hence the result.