Non-Equivalence

Theorem

 * $\neg \left ({p \iff q}\right) \dashv \vdash \left({\neg p \land q}\right) \lor \left({p \land \neg q}\right)$
 * $\neg \left ({p \iff q}\right) \dashv \vdash \neg \left({p \implies q}\right) \lor \neg \left({q \implies p}\right)$
 * $\neg \left ({p \iff q}\right) \dashv \vdash \left({p \lor q} \right) \land \neg \left({p \land q}\right)$
 * $\neg \left ({p \iff q}\right) \dashv \vdash \left({p \lor q} \right) \land \left({\neg p \lor \neg q}\right)$

Thus we see that negation of equivalence means the same thing as either-or but not both, that is, exclusive or.

Proof

 * align="right" | 2 ||
 * align="right" | 1
 * $\neg \left({\left ({p \implies q}\right) \land \left ({q \implies p}\right)}\right)$
 * Definition of Biconditional
 * 1
 * 1

The argument is reversible:


 * align="right" | 8 ||
 * align="right" | 1
 * $\neg \left ({p \iff q}\right)$
 * Definition of Biconditional
 * 7
 * 7


 * align="right" | 2 ||
 * align="right" | 1
 * $\neg \left({\left ({p \implies q}\right) \land \left ({q \implies p}\right)}\right)$
 * Definition of Biconditional
 * 1
 * 1

The above reasoning is completely reversible.


 * align="right" | 3 ||
 * align="right" | 1
 * $\neg \left ({p \iff q}\right)$
 * Definition of Biconditional
 * 2
 * 2


 * $\neg \left ({p \iff q}\right) \dashv \vdash \left({p \lor q} \right) \land \neg \left({p \land q}\right)$:

First, get this simple result:

... and its converse:

Now the main part of the proof:

The above argument is reversible:


 * $\neg \left ({p \iff q}\right) \dashv \vdash \left({p \lor q} \right) \land \left({\neg p \lor \neg q}\right)$

Follows directly from the above and De Morgan's Laws.

Proof by Truth Table
We apply the Method of Truth Tables to the propositions in turn.

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

$\begin{array}{|cccc||ccccccccc|} \hline \neg & (p & \iff & q) & (\neg & p & \land & q) & \lor & (p & \land & \neg & q) \\ \hline F & F & T & F & T & F & F & F & F & F & F & T & F \\ T & F & F & T & T & F & T & T & T & F & F & F & T \\ T & T & F & F & F & T & F & F & T & T & T & T & F \\ F & T & T & T & F & T & F & T & F & T & F & F & T \\ \hline \end{array}$

$\begin{array}{|cccc||ccccccccc|} \hline \neg & (p & \iff & q) & \neg & (p & \implies & q) & \lor & \neg & (q & \implies & p) \\ \hline F & F & T & F & F & F & T & F & F & F & F & T & F \\ T & F & F & T & F & F & T & T & T & T & T & F & F \\ T & T & F & F & T & T & F & F & T & F & F & T & T \\ F & T & T & T & F & T & T & T & F & F & T & T & T \\ \hline \end{array}$

$\begin{array}{|cccc||cccccccc|} \hline \neg & (p & \iff & q) & (p & \lor & q) & \land & \neg & (p & \land & q) \\ \hline F & F & T & F & F & F & F & F & T & F & F & F \\ T & F & F & T & F & T & T & T & T & F & F & T \\ T & T & F & F & T & T & F & T & T & T & F & F \\ F & T & T & T & T & T & T & F & F & T & T & T \\ \hline \end{array}$

$\begin{array}{|cccc||ccccccccc|} \hline \neg & (p & \iff & q) & (p & \lor & q) & \land & (\neg & p & \lor & \neg & q) \\ \hline F & F & T & F & F & F & F & F & T & F & T & T & F \\ T & F & F & T & F & T & T & T & T & F & T & F & T \\ T & T & F & F & T & T & F & T & F & T & T & T & F \\ F & T & T & T & T & T & T & F & F & T & F & F & T \\ \hline \end{array}$