# Rule of Material Equivalence/Formulation 1/Proof 1

## Theorem

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

## Proof

By the tableau method of natural deduction:

$p \iff q \vdash \left({p \implies q}\right) \land \left({q \implies p}\right)$
Line Pool Formula Rule Depends upon Notes
1 1 $p \iff q$ Premise (None)
2 1 $p \implies q$ Biconditional Elimination: $\iff \mathcal E_1$ 1
3 1 $q \implies p$ Biconditional Elimination: $\iff \mathcal E_2$ 1
4 1 $\left({p \implies q}\right) \land \left({q \implies p}\right)$ Rule of Conjunction: $\land \mathcal I$ 2, 3

By the tableau method of natural deduction:

$\left({p \implies q}\right) \land \left({q \implies p}\right) \vdash p \iff q$
Line Pool Formula Rule Depends upon Notes
1 1 $\left({p \implies q}\right) \land \left({q \implies p}\right)$ Premise (None)
2 1 $p \implies q$ Rule of Simplification: $\land \mathcal E_1$ 1
3 1 $q \implies p$ Rule of Simplification: $\land \mathcal E_2$ 1
4 1 $p \iff q$ Biconditional Introduction: $\iff \mathcal I$ 2, 3

$\blacksquare$