# Biconditional as Disjunction of Conjunctions/Formulation 2

## Theorem

$\vdash \paren {p \iff q} \iff \paren {\paren {p \land q} \lor \paren {\neg p \land \neg q} }$

## Proof 1

By the tableau method of natural deduction:

$\vdash \paren {p \iff q} \iff \paren {\paren {p \land q} \lor \paren {\neg p \land \neg q} }$
Line Pool Formula Rule Depends upon Notes
1 1 $p \iff q$ Assumption (None)
2 1 $\paren {p \land q} \lor \paren {\neg p \land \neg q}$ Sequent Introduction 1 Biconditional as Disjunction of Conjunctions: Formulation 1
3 $\paren {p \iff q} \implies \paren {\paren {p \land q} \lor \paren {\neg p \land \neg q} }$ Rule of Implication: $\implies \II$ 1 – 2 Assumption 1 has been discharged
4 4 $\paren {p \land q} \lor \paren {\neg p \land \neg q}$ Assumption (None)
5 4 $p \iff q$ Sequent Introduction 4 Biconditional as Disjunction of Conjunctions: Formulation 1
6 $\paren {\paren {p \land q} \lor \paren {\neg p \land \neg q} } \implies \paren {p \iff q}$ Rule of Implication: $\implies \II$ 4 – 5 Assumption 4 has been discharged
7 $\paren {p \iff q} \iff \paren {\paren {p \land q} \lor \paren {\neg p \land \neg q} }$ Biconditional Introduction: $\iff \II$ 3, 6

$\blacksquare$

## Proof by Truth Table

We apply the Method of Truth Tables.

As can be seen by inspection, in all cases the truth values under the main connective is true for all boolean interpretations.

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

$\blacksquare$