Biconditional is Transitive

Theorem
The biconditional operator is transitive:


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

This can otherwise be rendered as:


 * $\vdash \left({\left({p \iff q}\right) \land \left({q \iff r}\right)}\right) \implies \left({p \iff r}\right)$

These forms can be seen to be logically equivalent by application of the Extended Rule of Implication.

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

Proof by Truth Table
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.