Three-Way Exclusive Or and Equivalence

Theorem
Let $p \iff q$ be the biconditional operator, and $p \oplus q$ be the exclusive or operator.

Then:
 * $p \iff q \iff r \dashv \vdash p \oplus q \oplus r$

Proof
We apply the Method of Truth Tables to the proposition.

As can be seen by inspection, in each case, the truth values under the main connectives match for all models.

$\begin{array}{|ccccc||ccccc|} \hline (p & \iff & q) & \iff & r & (p & \oplus & q) & \oplus & r \\ \hline F & T & F & F & F & F & F & F & F & F \\ F & T & F & T & T & F & F & F & T & T \\ F & F & T & T & F & F & T & T & T & F \\ F & F & T & F & T & F & T & T & F & T \\ T & F & F & T & F & T & T & F & T & F \\ T & F & F & F & T & T & T & F & F & T \\ T & T & T & F & F & T & F & T & F & F \\ T & T & T & T & T & T & F & T & T & T \\ \hline \end{array}$

From Biconditional Properties and Exclusive Or Properties, we have that both $\iff$ and $\oplus$ are associative, which justifies the rendition of this result without parentheses.

Comment
A bizarrely non-intuitive result which is rarely documented.