p + q = (pVq) ^ -(p^q)
Here's how to show the first one, maybe you can figure out from this how to show the second. Truth tables are the conventional way to test whether two statements are logically equivalent. Look at the truth table below. You will notice that the blue columns are the same. One represents P + Q and the other represents (P v Q) ^ ~(P ^ Q). if two statements have the same truth values in every instant in a truth table, then they are logically equivalent
Recall that $\displaystyle p \Leftrightarrow q\quad \equiv \quad \left( { \sim p \wedge \sim q} \right) \vee \left( {p \wedge q} \right)$.
Thus
$\displaystyle \begin{array}{rcl}
\sim \left( {p + q} \right) & \equiv & \sim \left[ {\left( {p \vee q} \right) \wedge \sim \left( {p \wedge q} \right)} \right] \\
& \equiv & \left[ { \sim \left( {p \vee q} \right) \vee \left( {p \wedge q} \right)} \right] \\
\end{array}$
$\displaystyle \begin{array}{rcl}
\sim \left( {p + q} \right) & \equiv & \left[ {\left( { \sim p \wedge \sim q} \right) \vee \left( {p \wedge q} \right)} \right] \\
\; & \equiv & p \Leftrightarrow q \\
\end{array}$.
Thus
$\displaystyle \begin{array}{rcl}
\sim \left( {p + q} \right) & \equiv & \sim \left[ {\left( {p \vee q} \right) \wedge \sim \left( {p \wedge q} \right)} \right] \\
& \equiv & \left[ { \sim \left( {p \vee q} \right) \vee \left( {p \wedge q} \right)} \right] \\
\end{array}$