Non-Equivalence as Conjunction of Disjunction with Negation of Conjunction

Theorem

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

That is, negation of the biconditional means the same thing as either-or but not both, that is, exclusive or.

Proof 1

By the tableau method of natural deduction:

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

Line		Pool	Formula	Rule	Depends upon	Notes
1		1	$\neg \left ({p \iff q}\right)$	Premise	(None)
2		1	$\left({\neg p \land q}\right) \lor \left({p \land \neg q}\right)$	Sequent Introduction	1	Non-Equivalence as Disjunction of Conjunctions: Formulation 1
3		1	$\left({p \land \neg q}\right) \lor \left({\neg p \land q}\right)$	Sequent Introduction	2	Disjunction is Commutative
4		1	$\left({p \land \neg q}\right) \lor \left({q \land \neg p}\right)$	Sequent Introduction	3	Conjunction is Commutative
5		1	$\left ({\left({p \lor q}\right) \land \neg q}\right) \lor \left({\left({q \lor p}\right) \land \neg p}\right)$	Sequent Introduction	4	Conjunction of Disjunction with Negation is Conjunction with Negation
6		1	$\left ({\left({p \lor q}\right) \land \neg q}\right) \lor \left({\left({p \lor q}\right) \land \neg p}\right)$	Sequent Introduction	5	Disjunction is Commutative
7		1	$\left({p \lor q}\right) \land \left({\neg q \lor \neg p}\right)$	Sequent Introduction	6	Conjunction Distributes over Disjunction
8		1	$\left({p \lor q}\right) \land \left({\neg p \lor \neg q}\right)$	Sequent Introduction	7	Disjunction is Commutative
9		1	$\left({p \lor q}\right) \land \neg \left({\neg \neg p \land \neg \neg q}\right)$	Sequent Introduction	8	De Morgan's Laws: Disjunction
10		1	$\left({p \lor q}\right) \land \neg \left({p \land q}\right)$	Double Negation Elimination: $\neg \neg \mathcal E$	9

By the tableau method of natural deduction:

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

Line		Pool	Formula	Rule	Depends upon	Notes
1		1	$\left({p \lor q}\right) \land \neg \left({p \land q}\right)$	Premise	(None)
2		1	$\left({p \lor q}\right) \land \left({\neg p \lor \neg q}\right)$	Sequent Introduction	1	De Morgan's Laws: Disjunction of Negations
3		1	$\left({p \lor q}\right) \land \left({\neg p \lor \neg q}\right)$	Sequent Introduction	2	Disjunction is Commutative
4		1	$\left ({\left({p \lor q}\right) \land \neg q}\right) \lor \left({\left({p \lor q}\right) \land \neg p}\right)$	Sequent Introduction	3	Conjunction Distributes over Disjunction
5		1	$\left({p \land \neg q}\right) \lor \left({q \land \neg p}\right)$	Sequent Introduction	4	Conjunction of Disjunction with Negation is Conjunction with Negation
6		1	$\left({q \land \neg p}\right) \lor \left({p \land \neg q}\right)$	Sequent Introduction	5	Disjunction is Commutative
7		1	$\left({\neg p \land q}\right) \lor \left({p \land \neg q}\right)$	Sequent Introduction	6	Conjunction is Commutative
8		1	$\neg \left ({p \iff q}\right)$	Sequent Introduction	6	Non-Equivalence as Disjunction of Conjunctions: Formulation 1

$\blacksquare$

Proof 2

We apply the Method of Truth Tables.

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

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

$\blacksquare$