Biconditional is Reflexive

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:

$\vdash p \iff p$
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): $\S 2.3.3$
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