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:

... 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" | 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.

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

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

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

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