# Biconditional with Tautology

## Theorem

A biconditional with a tautology:

$p \iff \top \dashv \vdash p$

## Proof by Natural Deduction

By the tableau method of natural deduction:

$p \iff \top \vdash p$
Line Pool Formula Rule Depends upon Notes
1 1 $p \iff \top$ Premise (None)
2 $\top$ Rule of Top-Introduction: $\top \II$ (None)
3 1 $\top \implies p$ Biconditional Elimination: $\iff \EE_2$ 1
4 1 $p$ Modus Ponendo Ponens: $\implies \mathcal E$ 2, 3

$\Box$

By the tableau method of natural deduction:

$p \vdash p \iff \top$
Line Pool Formula Rule Depends upon Notes
1 1 $\top$ Premise (None)
2 2 $p$ Assumption (None)
3 $\top$ Rule of Top-Introduction: $\top \II$ (None)
4 $p \implies \top$ Rule of Implication: $\implies \II$ 2 – 3 Assumption 2 has been discharged
5 2 $\top \implies p$ Rule of Implication: $\implies \II$ 1 – 2 Assumption 1 has been discharged
6 2 $p \iff \top$ Biconditional Introduction: $\iff \II$ 4, 5

$\blacksquare$

## Proof by Truth Table

We apply the Method of Truth Tables to the proposition.

As can be seen by inspection, in each case, the truth values in the appropriate columns match for all boolean interpretations.

$\begin{array}{|c|ccc||c|ccc|} \hline p & p & \iff & \top & p \\ \hline F & F & F & T & F \\ T & T & T & T & T \\ \hline \end{array}$

$\blacksquare$