Properties of the deductive closure Let $\Phi_0$ be the set of $\cal L$-sentences. For $\Gamma\subseteq\Phi_0$, the deductive closure of $\Gamma$ is given by
$$\mathsf{Cn}(\Gamma)=\left\{\phi\in\Phi_0\mid\Gamma\vdash\phi\right\}$$
Typical properties of the consequence operator $\sf Cn$ are:


*

*$\Gamma\subseteq\mathsf{Cn}(\Gamma)$

*$\Sigma\subseteq\Gamma\Rightarrow\mathsf{Cn}(\Sigma)\subseteq\mathsf{Cn}(\Gamma)$

*$\mathsf{Cn}(\mathsf{Cn}(\Gamma))=\mathsf{Cn}(\Gamma)$


I understand what these properties mean, and could "argue" as to why they hold, but is there a formal way to prove them?
 A: I agree with most of @Max's answer, but I don't think you need the deduction theorem for item 3. Suppose you have a deduction $D$ of $\phi$ from some formulas $\psi_i\in C(\Gamma)$, so you also have deductions $D_i$ of each $\psi_i$ from $\Gamma$.  Take the deduction $D$ and, wherever any $\psi_i$ is used in $D$, replace it by its whole deduction $D_i$. (So you're inserting the $D_i$'s into $D$, to justify the formulas $\psi_i$ that were assumed in $D$.)  The result is a deduction of $\phi$ from $\Gamma$.
A: As you've indicated, the notion of deductive closure is defined relative to a notion of derivability (or consequence).  So, you can consider the three statements as really expressing properties of this or that notion of derivability.  Not every derivability notion has them all.  So, to verify that they hold, you'll need to consider specific properties of the system you're considering.  Depending on the system, this can be trivial or nontrivial.  
E.g., in a Hilbert-style system, all properties hold, and the verification is straightforward, given that the deduction theorem has been established. [Edit: as Andreas Blass points out, it's also straightforward without the deduction theorem!] 


*

*suppose $\phi$ is in $\Gamma$.  Then $\phi$ itself is a proof of $\phi$ from $\Gamma$.  So $\Gamma\subseteq \text{Cn}(\Gamma)$.

*suppose $\phi$ is in $\text{Cn}(\Sigma)$, so that, say, $P$ is a proof of $\phi$ from $\Sigma$. If $\Sigma\subseteq \Gamma$, then $P$ itself is a proof of $P$ from $\Gamma$.  So $\text{Cn}(\Sigma)\subseteq\text{Cn}(\Gamma)$.

*suppose $\phi$ is in $\text{Cn}(\text{Cn}(\Gamma))$.  Then there is a proof of $\phi$ from $\text{Cn}(\Gamma)$.  This proof must use only finitely many axioms from $\text{Cn}(\Gamma)$, say $\psi_1,\ldots,\psi_k$.  By the deduction theorem, it follows that there is a proof $Q$ of
$\psi_1\to\cdots\to\psi_k\to \phi$
from the empty set.  However, since $\psi_1,\ldots,\psi_k$ are in $\text{Cn}(\Gamma)$, therefore there are proofs $Q_1,\ldots,Q_k$ of each from $\Gamma$ itself.  After concatenating $Q_1,\ldots,Q_k$ to $Q$, the result of appending $k$ applications of modus ponens is a proof of $\phi$ from $\Gamma$.  So $\text{Cn}(\text{Cn}(\Gamma))\subseteq \text{Cn}(\Gamma)$.  The converse follows from 1.
