# Biconditional iff Disjunction implies Conjunction/Formulation 1/Proof

## Theorem

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

## Proof

We apply the Method of Truth Tables.

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

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

$\blacksquare$