Biconditional is Transitive/Formulation 1
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:
| 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: $\iff \EE_1$ | 1 | ||
| 4 | 2 | $q \implies r$ | Biconditional Elimination: $\iff \EE_1$ | 2 | ||
| 5 | 1, 2 | $p \implies r$ | Sequent Introduction | 1, 2 | Hypothetical Syllogism: Formulation 1 | |
| 6 | 1 | $q \implies p$ | Biconditional Elimination: $\iff \EE_2$ | 1 | ||
| 7 | 2 | $r \implies q$ | Biconditional Elimination: $\iff \EE_2$ | 2 | ||
| 8 | 1, 2 | $r \implies p$ | Sequent Introduction | 7, 6 | Hypothetical Syllogism: Formulation 1 | |
| 9 | 1, 2 | $p \iff r$ | Biconditional Introduction: $\iff \II$ | 5, 8 |
$\blacksquare$
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.
$\blacksquare$
Sources
- 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 1$. Sets; inclusion; intersection; union; complementation; number systems
- 1996: H. Jerome Keisler and Joel Robbin: Mathematical Logic and Computability ... (previous) ... (next): $\S 1.13$: Tableau Problems (TAB1): EQUIV