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
 * $\neg \left({\left ({p \implies q}\right) \land \left ({q \implies p}\right)}\right)$
 * Material Equivalence
 * 1
 * align="right" | 3 ||
 * align="right" | 1
 * $\neg \left({\left ({\neg p \lor q}\right) \land \left ({\neg q \lor p}\right)}\right)$
 * Rule of Material Implication
 * 2
 * align="right" | 4 ||
 * align="right" | 1
 * $\neg \left({\neg p \lor q}\right) \lor \neg \left ({\neg q \lor p}\right)$
 * DM
 * 3
 * align="right" | 5 ||
 * align="right" | 1
 * $\left ({\neg \neg p \land \neg q}\right) \lor \left ({\neg \neg q \land \neg p}\right)$
 * DM
 * 4
 * align="right" | 6 ||
 * align="right" | 1
 * $\left ({p \land \neg q}\right) \lor \left ({q \land \neg p}\right)$
 * $\neg \neg \mathcal{E}$
 * 5
 * align="right" | 7 ||
 * align="right" | 1
 * $\left ({p \land \neg q}\right) \lor \left ({\neg p \land q}\right)$
 * Comm
 * 6
 * align="right" | 8 ||
 * align="right" | 1
 * $\left ({\neg p \land q}\right) \lor \left ({p \land \neg q}\right)$
 * Comm
 * 7
 * align="right" | 7 ||
 * align="right" | 1
 * $\left ({p \land \neg q}\right) \lor \left ({\neg p \land q}\right)$
 * Comm
 * 6
 * align="right" | 8 ||
 * align="right" | 1
 * $\left ({\neg p \land q}\right) \lor \left ({p \land \neg q}\right)$
 * Comm
 * 7
 * $\left ({\neg p \land q}\right) \lor \left ({p \land \neg q}\right)$
 * Comm
 * 7

The argument is reversible:


 * align="right" | 2 ||
 * align="right" | 1
 * $\left ({p \land \neg q}\right) \lor \left ({\neg p \land q}\right)$
 * Comm
 * 1
 * align="right" | 3 ||
 * align="right" | 1
 * $\left ({p \land \neg q}\right) \lor \left ({q \land \neg p}\right)$
 * Comm
 * 2
 * align="right" | 4 ||
 * align="right" | 1
 * $\left ({\neg \neg p \land \neg q}\right) \lor \left ({\neg \neg q \land \neg p}\right)$
 * $\neg \neg \mathcal{E}$
 * 3
 * align="right" | 5 ||
 * align="right" | 1
 * $\neg \left({\neg p \lor q}\right) \lor \neg \left ({\neg q \lor p}\right)$
 * DM
 * 4
 * align="right" | 6 ||
 * align="right" | 1
 * $\neg \left({\left ({\neg p \lor q}\right) \land \left ({\neg q \lor p}\right)}\right)$
 * DM
 * 5
 * align="right" | 7 ||
 * align="right" | 1
 * $\neg \left({\left ({p \implies q}\right) \land \left ({q \implies p}\right)}\right)$
 * Rule of Material Implication
 * 6
 * align="right" | 8 ||
 * align="right" | 1
 * $\neg \left ({p \iff q}\right)$
 * Material Equivalence
 * 7
 * align="right" | 7 ||
 * align="right" | 1
 * $\neg \left({\left ({p \implies q}\right) \land \left ({q \implies p}\right)}\right)$
 * Rule of Material Implication
 * 6
 * align="right" | 8 ||
 * align="right" | 1
 * $\neg \left ({p \iff q}\right)$
 * Material Equivalence
 * 7
 * $\neg \left ({p \iff q}\right)$
 * Material Equivalence
 * 7


 * align="right" | 2 ||
 * align="right" | 1
 * $\neg \left({\left ({p \implies q}\right) \land \left ({q \implies p}\right)}\right)$
 * Material Equivalence
 * 1
 * align="right" | 3 ||
 * align="right" | 1
 * $\neg \left({p \implies q}\right) \lor \neg \left({q \implies p}\right)$
 * DM
 * 2
 * $\neg \left({p \implies q}\right) \lor \neg \left({q \implies p}\right)$
 * DM
 * 2

The above reasoning is completely reversible.


 * align="right" | 2 ||
 * align="right" | 1
 * $\neg \left({\left ({p \implies q}\right) \land \left ({q \implies p}\right)}\right)$
 * DM
 * 1
 * align="right" | 3 ||
 * align="right" | 1
 * $\neg \left ({p \iff q}\right)$
 * Material Equivalence
 * 2
 * $\neg \left ({p \iff q}\right)$
 * Material Equivalence
 * 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:


 * align="right" | 3 ||
 * align="right" | 1
 * $p \lor q$
 * $\lor \mathcal{I}_1$
 * 2
 * 2

... and its converse:


 * align="right" | 2 ||
 * align="right" | 1
 * $\left({p \land \neg q}\right) \lor \left({q \land \neg q}\right)$
 * Dist
 * 1
 * 1


 * align="right" | 4 ||
 * align="right" | 3
 * $\bot$
 * $\neg \mathcal{E}$
 * 3
 * 3


 * align="right" | 6 ||
 * align="right" | 1
 * $p \land \neg q$
 * Disjunctive Syllogism
 * 2, 5
 * 2, 5

Now the main part of the proof:


 * align="right" | 2 ||
 * align="right" | 1
 * $\left({\neg p \land q}\right) \lor \left({p \land \neg q}\right)$
 * Sequent Introduction
 * 1
 * Non-Equivalence: from above
 * align="right" | 3 ||
 * align="right" | 1
 * $\left({p \land \neg q}\right) \lor \left({q \land \neg p}\right)$
 * Comm
 * 2
 * align="right" | 4 ||
 * align="right" | 1
 * $\left ({\left({p \lor q}\right) \land \neg q}\right) \lor \left({\left({q \lor p}\right) \land \neg p}\right)$
 * Sequent Introduction
 * 3
 * From above
 * align="right" | 5 ||
 * align="right" | 1
 * $\left ({\left({p \lor q}\right) \land \neg q}\right) \lor \left({\left({p \lor q}\right) \land \neg p}\right)$
 * Comm
 * 4
 * align="right" | 6 ||
 * align="right" | 1
 * $\left({p \lor q}\right) \land \left({\neg q \lor \neg p}\right)$
 * Dist
 * 5
 * align="right" | 7 ||
 * align="right" | 1
 * $\left({p \lor q}\right) \land \left({\neg p \lor \neg q}\right)$
 * Comm
 * 6
 * align="right" | 8 ||
 * align="right" | 1
 * $\left({p \lor q}\right) \land \neg \left({\neg \neg p \land \neg \neg q}\right)$
 * DM
 * 7
 * align="right" | 9 ||
 * align="right" | 1
 * $\left({p \lor q}\right) \land \neg \left({p \land q}\right)$
 * $\neg \neg \mathcal{E}$
 * 8
 * align="right" | 8 ||
 * align="right" | 1
 * $\left({p \lor q}\right) \land \neg \left({\neg \neg p \land \neg \neg q}\right)$
 * DM
 * 7
 * align="right" | 9 ||
 * align="right" | 1
 * $\left({p \lor q}\right) \land \neg \left({p \land q}\right)$
 * $\neg \neg \mathcal{E}$
 * 8
 * $\left({p \lor q}\right) \land \neg \left({p \land q}\right)$
 * $\neg \neg \mathcal{E}$
 * 8

The above argument is reversible:


 * align="right" | 2 ||
 * align="right" | 1
 * $\left({p \lor q}\right) \land \left({\neg p \lor \neg q}\right)$
 * DM
 * 1
 * align="right" | 3 ||
 * align="right" | 1
 * $\left({p \lor q}\right) \land \left({\neg q \lor \neg p}\right)$
 * Comm
 * 2
 * align="right" | 4 ||
 * align="right" | 1
 * $\left ({\left({p \lor q}\right) \land \neg q}\right) \lor \left({\left({p \lor q}\right) \land \neg p}\right)$
 * Dist
 * 3
 * align="right" | 5 ||
 * align="right" | 1
 * $\left({p \land \neg q}\right) \lor \left({q \land \neg p}\right)$
 * SI
 * 4
 * From above
 * align="right" | 6 ||
 * align="right" | 1
 * $\left({\neg p \land q}\right) \lor \left({p \land \neg q}\right)$
 * Comm
 * 5
 * align="right" | 7 ||
 * align="right" | 1
 * $\neg \left ({p \iff q}\right)$
 * SI
 * 6
 * Non-Equivalence: from above
 * Comm
 * 5
 * align="right" | 7 ||
 * align="right" | 1
 * $\neg \left ({p \iff q}\right)$
 * SI
 * 6
 * Non-Equivalence: from above
 * 6
 * Non-Equivalence: from above


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

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.

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

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

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

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