Double Negation/Formulation 1/Proof 2

Theorem

 * $p \dashv \vdash \neg \neg p$

Proof
We apply the Method of Truth Tables to the proposition.

As can be seen by inspection, appropriate truth values match for both models.

$\begin{array}{|c||ccc|} \hline p & \neg & \neg & p \\ \hline F & F & T & F \\ T & T & F & T \\ \hline \end{array}$

Hence $p \dashv \vdash \neg \neg p$.