Biconditional is Reflexive
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Theorem
The biconditional operator, considered as a relation, is reflexive:
- $\vdash p \iff p$
This can otherwise be stated as that equivalence destroys copies of itself.
Proof 1
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | $p \implies p$ | Theorem Introduction | (None) | Law of Identity: Formulation 2 | ||
2 | $p \iff p$ | Biconditional Introduction: $\iff \II$ | 1, 1 |
$\blacksquare$
Proof 2
We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connective match for both boolean interpretations.
$\begin{array}{|ccc|} \hline p & \iff & p \\ \hline F & T & F \\ T & T & T \\ \hline \end{array}$
$\blacksquare$
Sources
- 1959: A.H. Basson and D.J. O'Connor: Introduction to Symbolic Logic (3rd ed.) ... (previous) ... (next): $\S 4.7$: The Derivation of Formulae: $D \, 23$
- 1964: Donald Kalish and Richard Montague: Logic: Techniques of Formal Reasoning ... (previous) ... (next): $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 5$: Theorem $\text{T91}$
- 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 1$. Sets; inclusion; intersection; union; complementation; number systems
- 2012: M. Ben-Ari: Mathematical Logic for Computer Science (3rd ed.) ... (previous) ... (next): $\S 2.3.3$