Biconditional is Commutative/Formulation 1

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Theorems

The biconditional operator is commutative:

$p \iff q \dashv \vdash q \iff p$


Proof 1

By the tableau method of natural deduction:

$p \iff q \vdash q \iff p$
Line Pool Formula Rule Depends upon Notes
1 1 $p \iff q$ Premise (None)
2 1 $p \implies q$ Biconditional Elimination: $\iff \EE_1$ 1
3 1 $q \implies p$ Biconditional Elimination: $\iff \EE_2$ 1
4 1 $q \iff p$ Biconditional Introduction: $\iff \II$ 3, 2

$\Box$


By the tableau method of natural deduction:

$q \iff p \vdash p \iff q$
Line Pool Formula Rule Depends upon Notes
1 1 $q \iff p$ Premise (None)
2 1 $q \implies p$ Biconditional Elimination: $\iff \EE_1$ 1
3 1 $p \implies q$ Biconditional Elimination: $\iff \EE_2$ 1
4 1 $p \iff q$ Biconditional Introduction: $\iff \II$ 3, 2

$\blacksquare$


Proof 2

By the tableau method of natural deduction:

$p \iff q \vdash q \iff p$
Line Pool Formula Rule Depends upon Notes
1 1 $p \iff q$ Premise (None)
2 1 $\paren {p \implies q} \land \paren {q \implies p}$ Sequent Introduction 1 Rule of Material Equivalence
3 1 $\paren {q \implies p} \land \paren {p \implies q}$ Sequent Introduction 2 Conjunction is Commutative
4 1 $q \iff p$ Sequent Introduction 3 Rule of Material Equivalence

$\Box$


By the tableau method of natural deduction:

$q \iff p \vdash p \iff q$
Line Pool Formula Rule Depends upon Notes
1 1 $q \iff p$ Premise (None)
2 1 $\paren {q \implies p} \land \paren {p \implies q}$ Sequent Introduction 1 Rule of Material Equivalence
3 1 $\paren {p \implies q} \land \paren {q \implies p}$ Sequent Introduction 2 Conjunction is Commutative
4 1 $p \iff q$ Sequent Introduction 3 Rule of Material Equivalence

$\blacksquare$


Proof by Truth Table

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.


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

$\blacksquare$


Sources

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