Biconditional is Transitive/Formulation 1

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Theorem

The biconditional operator is transitive:

$p \iff q, q \iff r \vdash p \iff r$


Proof 1

By the tableau method of natural deduction:

$p \iff q, q \iff r \vdash p \iff r$
Line Pool Formula Rule Depends upon Notes
1 1 $p \iff q$ Premise (None)
2 2 $q \iff r$ Premise (None)
3 1 $p \implies q$ Biconditional Elimination: $\iff \mathcal E_1$ 1
4 2 $q \implies r$ Biconditional Elimination: $\iff \mathcal E_1$ 2
5 1, 2 $p \implies r$ Sequent Introduction 1, 2 Hypothetical Syllogism: Formulation 1
6 1 $q \implies p$ Biconditional Elimination: $\iff \mathcal E_2$ 1
7 2 $r \implies q$ Biconditional Elimination: $\iff \mathcal E_2$ 2
8 1, 2 $r \implies p$ Sequent Introduction 7, 6 Hypothetical Syllogism: Formulation 1
9 1, 2 $p \iff r$ Biconditional Introduction: $\iff \mathcal I$ 5, 8


Proof 2

We apply the Method of Truth Tables.

As can be seen for all boolean interpretations by inspection, where the truth values under the main connective on the left hand side is $T$, that under the one on the right hand side is also $T$:

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

Hence the result.

$\blacksquare$


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