Non-Equivalence

Non-Equivalence as Disjunction of Conjunctions

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

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

Proof
The above reasoning is completely reversible.


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

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.

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

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

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

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