Proving if $\varphi(c) \in \Sigma$ then $\exists x \, \varphi(x) \in \Sigma$ for a maximal consistent set

Question

Let's say $\varphi$ is a formula and $\Sigma$ is a set of formulas. Let's then say that $\Sigma$ is a maximal set. All I know about maximal sets is that either $\varphi \in \Sigma$ or $\lnot\varphi \in \Sigma$ (where $\varphi$ is a formula). So using that definition, if $c$ is a constant then how can I prove that if $\varphi(c) \in \Sigma$ then $\exists x \, \varphi(x) \in \Sigma$ if $\Sigma$ is a maximal consistent set (consistent just means the formulas in $\Sigma$ will never lead to a contradiction in a derivation).

Answer 1

Proof by contradiction. If, by absurd, $\exists x \, \varphi(x) \notin \Sigma$, then $\lnot \exists x \, \varphi(x) \in \Sigma$ by maximality of $\Sigma$. But then $\Sigma$ would not be consistent, because from $\varphi(c)$ and $\lnot\exists x \, \varphi(x)$ you can easily derive a contradiction, as the following derivation in natural deduction shows.

\begin{align} \dfrac{\lnot\exists x \, \varphi(x) \qquad \dfrac{\varphi(c)}{\exists x \, \varphi(x)}\exists_\text{intro}}{\bot}\lnot_\text{elim} \end{align}

Therefore $\exists x \, \varphi(x) \in \Sigma$, since $\Sigma$ is consistent.