Fiction variables? In every Boolean function $f(x_1, x_2,\ldots, x_n)$, for every $i$  ($1\le i\le n$), $x_i$ is called fiction variable if and only if when for every Boolean assessment for the rest variables $x_1, x_2,\ldots, x_{(i-1)}, x_{(i+1)},\ldots,x_n$ the function has the same value for $x_i=0$ and $x_i=1$.
How many Boolean functions of $n$ variables have no fictional variables?
I have some ideas, but don't know how to prove it "formally". Any help would be greatly appreciated. 
 A: I think the question is how many Boolean functions of $n$ variables have no fictional variables.  This can be answered using the inclusion-exclusion principle.  If you take a subset $S$ of size $k$ of the $n$ variables and ask how many Boolean functions on $n$ variables have at least the variables in $S$ fictional, the answer is the same as the number of Boolean functions on the $n-k$ remaining variables, which is $2^{2^{n-k}}.$  Therefore, the number of Boolean functions on $n$ variables with no fictional variables is 
$$
\sum_{0\le k\le n} (-1)^k \binom{n}{k} 2^{2^{n-k}}.\qquad (*)
$$
This is OEIS A000371.
In response to the comment: the inclusion-exclusion principle says that if we have a finite set $U$ of properties and a set of objects, and if the number of objects with at least the properties in the set $T\subseteq U$ is $N(T)$, then the number of objects with none of the properties is
$$
N_0=\sum_{T\subseteq U} (-1)^{|T|} N(T). \qquad (+)
$$
(*) follows immediately from (+) upon letting $\{x_1, \dots, x_n\}$ be the set of variables and letting $U$ be the set {"has at least $x_1$ as a fictional variable", $\dots$, "has at least $x_n$ as a fictional variable"}.  This is because, if $T$ is a subset of $U$ of size $k$, then, as explained above, the set of Boolean functions on $n$ variables with at least the properties in $T$ has size $N(T)=2^{2^{n-k}}$.
To see why (+) is true, take as an example the case where $U$ has size 3.  Then, if we let $X$ be the set of all objects and $A$, $B$ and $C$ be the subsets of $X$ which satisfy the first, second, and third properties of $U$, (+) reduces to
$$
|X\setminus (A\cup B\cup C)|=
$$
$$|X|-|A|-|B|-|C|+|A\cap B|+|A\cap C|+|B\cap C|-|A\cap B\cap C|.\ \  (++)
$$
On the right-hand side of (++), to count the number of objects with no properties, we start with the number of all objects, $|X|$, which is an overestimate.  We then subtract $|A|+|B|+|C|$.  This produces an underestimate, since
$|X|-|A|-|B|-|C|$ correctly counts each object with no properties once and each object with exactly one property $0$ times, but if an object has exactly two properties, it is counted $-1$ times, and if it has all three properties, it is counted $-2$ times.  To remedy this, we add
$|A\cap B|+|A\cap C|+|B\cap C|$.  This results in a count which is correct for objects with no properties, counting them once, or objects with exactly $1$ or $2$ properties, which are counted $0$ times.  However, it overcounts objects with all three properties by counting them once instead of $0$ times.  This is fixed by subtracting $|A\cap B\cap C|$, giving the correct and final form of (++).  
The proof of (+) in the general case is similar.  The inclusion-exclusion principle is also often stated in the form
$$
N_{\ge 1}=N(\emptyset)-N_0=\sum_{\emptyset\ne T\subseteq U} (-1)^{|T|-1} N(T),
$$
where $N_{\ge 1}$ is now the total number of objects with at least one property.
