Biconditional as Disjunction of Conjunctions/Formulation 2/Proof 1

Theorem

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

Proof

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: $\implies \mathcal I$ 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: $\implies \mathcal I$ 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: $\iff \mathcal I$ 3, 6

$\blacksquare$