Problem in understanding Graph Enumeration ? (From the book on Graphical Enumeration by Palmer and Harary) I was reading the first answer to this question, where the answer refers to a book by F. Harary and E.M. Palmer. In the book on page 8, they provide the following result:

Theorem: The exponential generating functions $G(x)$ and $C(x)$ for labeled
graphs and labeled connected graphs come to terms in the following relation
$$1+G(x) = e^{C(x)}$$

What is intriguing to me is the following statement (the last paragraph of page 8) :

Furthermore, it is evident that if the exponential generating function for a
class of graphs is known, then the exponential generating function for the
corresponding connected graphs will be the formal logarithm of the first
series, just as in (1 .2.6) for all graphs

1.2.6 is the equation above.
So if I have a generating function, $G'(x)$, for any arbitrary property,$P(a)$, on labelled graphs, then I can find generating function for a labelled graph which obeys $P(a)$ and is connected as follows:
$$1+G'(x) = e^{C'(x)}$$
Where $C'(x)$ is the generating function for the connected graph such that it obeys $P(a)$.
Is my understanding of the statement correct ?
So for example if $G'(x)$ is the generating function for 2-regular graphs then the generating function for $C'(x)$ i.e connected 2-regular graphs is given as follows :
$$C'(x) = \log(1+G'(x))$$
 A: With this question I would cast it in a more modern light, namely
combinatorial classes as introduced by Flajolet and Sedgewick in
Analytic Combinatorics. They define the set operator applied to a
combinatorial class $\mathcal{C}$ as follows:
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}
\mathcal{G} = \textsc{SET}(\mathcal{C})
= \sum_{k\ge 0} \textsc{SET}_{=k}(\mathcal{C}).$$
Note that this operator includes $\textsc{SET}_{=0}(\mathcal{C})$
which is the empty set of size zero.
This gives the generating function for $\textsc{SET}(\mathcal{C})$
$$\sum_{k\ge 0} \frac{C(z)^k}{k!} = \exp C(z).$$
Now $\mathcal{G}$ and $\mathcal{C}$ are in a set-of relation but
$\mathcal{G}$ may or may not include the empty graph. In the latter case
we get instead of $\textsc{SET}(\mathcal{C})$ the class
$\textsc{SET}_{\ge 1}(\mathcal{C})$ with generating function
$$-1 + \exp C(z).$$
It follows that we have two cases, if $\mathcal{G}$ includes the empty
graph we get
$$G(z) = \exp C(z)$$
and if it does not,
$$1 + G(z) = \exp C(z).$$
In your example of $2$-regular graph the EGF is constructed from the
set operator $\textsc{SET}$ as documented e.g. at this MSE
link and hence it
produces the empty graph as belonging to the class of graphs (a graph  on
no vertices with no edges does indeed have all vertices of degree two)
and hence the correction term of one plus is not needed and we find
$$C(z) = \log G(z).$$
We can check this by computing the constant term of
$\exp(-z/2-z^2/4)/\sqrt{1-z}$ which is one. 
Another example where the correction term is not needed is the class of
connected labeled graphs with some number of edges, which was  discussed
as this MSE link  II.
It should be clear that connectivity induces a set-of relationship as
we may permute the connected components.
