# Biconditional iff Disjunction implies Conjunction/Formulation 1/Forward Implication

## Theorem

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

## Proof

By the tableau method of natural deduction:

$p \iff q \vdash \left({p \lor q}\right) \implies \left({p \land q}\right)$
Line Pool Formula Rule Depends upon Notes
1 1 $p \iff q$ Premise (None)
2 1 $\left({p \land q}\right) \lor \left({\neg p \land \neg q}\right)$ Sequent Introduction 1 Biconditional as Disjunction of Conjunctions
3 1 $\left({p \land q}\right) \lor \neg \left({p \lor q}\right)$ Sequent Introduction 2 De Morgan's Laws: Conjunction of Negations
4 1 $\neg \left({p \lor q}\right) \lor \left({p \land q}\right)$ Sequent Introduction 3 Disjunction is Commutative
5 1 $\left({p \lor q}\right) \implies \left({p \land q}\right)$ Sequent Introduction 4 Rule of Material Implication

$\blacksquare$