Properties of Biconditional

Theorem

 * $$p \iff q \dashv \vdash \left({p \and q}\right) \or \left({\neg p \and \neg q}\right)$$
 * $$p \iff q \dashv \vdash \neg p \iff \neg q$$
 * $$p \iff q \dashv \vdash \left({p \or q}\right) \implies \left({p \and q}\right)$$

Proof by Natural Deduction
By the tableau method:

The argument reverses:

The argument reverses:

Proof by Truth Table
We apply the Method of Truth Tables to the propositions in turn.

As can be seen by inspection, in all cases the truth values under the main connectives match for all models.

$$\begin{array}{|ccc||ccccccccc|} \hline p & \iff & q & (p & \and & q) & \or & (\neg & p & \and & \neg & q) \\ \hline F & T & F & F & F & F & T & T & F & T & T & F \\ F & F & T & F & F & T & F & T & F & F & F & T \\ T & F & F & T & F & F & F & F & T & F & T & F \\ T & T & T & T & T & T & T & F & T & F & F & T \\ \hline \end{array}$$

$$\begin{array}{|ccc||ccccc|} \hline p & \iff & q & \neg & p & \iff & \neg & q \\ \hline F & T & F & T & F & T & T & F \\ F & F & T & T & F & F & F & T \\ T & F & F & F & T & F & T & F \\ T & T & T & F & T & T & F & T \\ \hline \end{array}$$

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