# 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.

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

$\blacksquare$