I am trying to find the shortest equivalent expression of the following:
((C → D) $\wedge$ (D → C)) $↔$ (C $\wedge$ D ∨ ¬C $\wedge$ ¬D)
I have "simplified" the expression into the following:
(($\neg$C $\vee$ D) $\wedge$ ($\neg$D $\vee$ C)) $↔$ ((C $\wedge$ D) ∨ (¬C $\wedge$ ¬D))
I am able to figure our that the expression on the left hand side of the $↔$ operator requires either D or C to be true to evaluate to true. However, the expression on the right hand side requires both to be either true or both to be false.
However I am not sure on how to find an even shorter equivalent expression than I already have. Also, would the resulting expression be a tautology or contradiction?
Your simplification is correct. One key quality of the material conditional(if) in classical logic is that $C \rightarrow D$ is equivalent to $\lnot C \lor D$. You can read more about the material conditional here Since your simplication is correct it can be shown that the statement is a tautology. Use De Morgan's law to simplify the second portion and you will see \begin{equation} (C \land D) \lor (\lnot C \land \lnot D) \end{equation} which is equivalent to \begin{equation} (C \land D) \lor \lnot(C \lor D) \end{equation} You can see that this is only true if C and D are both true or both false. This is quivalent to the not exclusive or (XNOR) which is also equivalent to the if and only if condition. Therefore the statement becomes \begin{equation} (\lnot C \lor D) \land (C \lor \lnot D) \leftrightarrow (C \leftrightarrow D) \end{equation} But you can also see that on the left side both C and D need to both be true or both be false. This is just the XNOR which is just the if and only if. The final expression of your statment is \begin{equation} (C \leftrightarrow D) \leftrightarrow (C \leftrightarrow D) \end{equation} Which is always true and therefore a tautology.