Rule of Material Equivalence/Formulation 1

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Theorem

The biconditional operation can be interpreted as the conjunction of implications:

$p \iff q \dashv \vdash \paren {p \implies q} \land \paren {q \implies p}$


Proof 1

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

$\blacksquare$


Proof 2

We apply the Method of Truth Tables.

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

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

$\blacksquare$


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