Biconditional is Transitive/Formulation 1

Theorem
The biconditional operator is transitive:
 * $p \iff q, q \iff r \vdash p \iff r$

Proof
Proof by natural deduction would be more tedious than illuminating.

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.