\emph{A formula $\varphi$ is valid, if and only if, $\neg \varphi$ is not satisfiable}.

Consider the formula $\varphi=(x \vee \neg x)$. This formula is valid, i.e., all rows in the truth table 
would evaluate to \emph{true}. The negation of $\varphi$ is the following: 
$\neg \varphi = \neg (x \vee \neg x) = \neg x \wedge x$, which is not satisfiable, i.e., all rows the truth table would evaluate to \emph{false}.

\noindent\emph{A formula $\varphi$ is satisfiable, if and only if, $\neg \varphi$ is not valid}.

If $\varphi$ is satisfiable, there is at least one model that makes the formula true. If we negate the formula,
these models make the negated formula false, and therefore, the negated formula cannot be valid.
Consider the formula $\varphi=(x \vee y)$.
There is at least one model that makes the formula true, e.g. $\mathcal{M} \coloneqq x = T, \ y = T$.
The negation of $\varphi$ is the following: 
$\neg \varphi = \neg (x \vee y) = \neg x \wedge \neg y$.
Under the same model $\mathcal{M}$ as before, $\neg \varphi$ evaluates to false.
So the negated formula is not valid.
