Rule of Material Equivalence/Formulation 1/Proof by Truth Table

Theorem

$p \iff q \dashv \vdash \paren {p \implies q} \land \paren {q \implies p}$

Proof

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.

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

$\blacksquare$

Sources

• 1959: A.H. Basson and D.J. O'Connor: Introduction to Symbolic Logic (3rd ed.) ... (previous) ... (next): $\S 2.5$: Further Logical Constants
• 1971: Robert H. Kasriel: Undergraduate Topology ... (previous) ... (next): $\S1.2$: Some Remarks on the Use of the Connectives and, or, implies
• 1980: D.J. O'Connor and Betty Powell: Elementary Logic ... (previous) ... (next): $\S \text{I}: 12$: Material Equivalence and Alternation