Many people (including mathematicians) often mistakenly use $ P \Leftrightarrow Q \Leftrightarrow R $ to indicate equivalence, that is $ (P \Leftrightarrow Q) \wedge (Q \Leftrightarrow R) $, which is a way of saying $ P $ is equivalent to $ Q $ which is also equivalent to $ R $. (See goblin's answer to the question “IFF” (if and only if) vs. “TFAE” (the following are equivalent).)
I think it is confusing to think of if-and-only-if as equivalence relationship, since being an associative operator(1), we are allowed to write expressions such as $ P \Leftrightarrow Q \Leftrightarrow R $, which are not ambiguous at all (compare with $ P \Rightarrow Q \Rightarrow R $, which is only equivalent to $ P \Rightarrow (Q \Rightarrow R) $).
lhf's solution works fine if you are trying to establish an equivalence relationship between only two sub-expressions, but XNOR is not an equivalent of iff operator! So, for instance, compare the following two expressions: ($ \bar\veebar $ is used for XNOR)
$ \begin{array}{|c|c|c|c|c|} \hline P & Q & R & P \Leftrightarrow Q \Leftrightarrow R & P\ \bar\veebar\ Q\ \bar\veebar\ R \\ \hline \top & \top & \top & \top & \bot \\ \hline \top & \top & \bot & \bot & \top \\ \hline \top & \bot & \top & \bot & \top \\ \hline \top & \bot & \bot & \top & \bot \\ \hline \bot & \top & \top & \bot & \top \\ \hline \bot & \top & \bot & \top & \bot \\ \hline \bot & \bot & \top & \top & \bot \\ \hline \bot & \bot & \bot & \bot & \top \\ \hline \end{array} $
As can be seen in the truth table, XNOR and iff operators are not equivalent; furthermore, you might have noticed that they are negation of each other!
I am going to omit the proof, simply because I don't have one, but a given
$ P_1 \Leftrightarrow P_2 \Leftrightarrow \ldots \Leftrightarrow P_n $
is equivalent to
$ P_1\ \bar\veebar\ P_2\ \bar\veebar\ \ldots\ \bar\veebar\ P_n $
if and only if $ n $ is an even number; else is equivalent to
$ P_1 \veebar P_2 \veebar \ldots \veebar P_n $
where $ \veebar $ is used for XOR.
Both XNOR()
and XOR()
operators are available on WolframAlpha.
(1): $ P \Leftrightarrow Q \Leftrightarrow R $ is same as $ (P \Leftrightarrow Q) \Leftrightarrow R $ which is also same as $ P \Leftrightarrow (Q \Leftrightarrow R) $.