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 boolean interpretations.

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

$\blacksquare$

From Biconditional is Associative and Exclusive Or is Associative, 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.