# 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:

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:

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:

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:

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

- 2012: M. Ben-Ari:
*Mathematical Logic for Computer Science*(3rd ed.) ... (previous) ... (next): $\S 2.3.3$