Describe all irreducible voting systems with two outcomes A situation where $n$ voters choose between two candidates can be modelized
by a $n$-uple $(a_1,\ldots,a_n)\in\lbrace 0,1 \rbrace^n$, where $a_i$ denotes
the wish of the $i$-th voter.  So the voting system (the rule which decides
which candidate will be elected from the uple) can be described as a boolean
map $f:\lbrace 0,1 \rbrace^n \to \lbrace 0,1 \rbrace$. The following assumptions 
are natural on $f$ :
(1) Symmetry : $f(1-a_1,1-a_2,\ldots,1-a_n)=1-f(a_1,\ldots,a_n)$ for any uple
$a_1,\ldots,a_n$.
(2) Monotonicity : if $a_k\leq b_k$ for every $k$, then 
$f(a_1,\ldots,a_n) \leq f(b_1,\ldots,b_n)$.
If $I\subsetneq \lbrace 1,2,\ldots, n \rbrace$ is such that $f$ does not depend on the coordinates $a_j$ for $j\not\in I$, we say that $I$ is a (nontrivial) coalition ; the trivial coalition is $I=\lbrace 1,2,\ldots, n \rbrace$. When there
is no nontrivial coalition, we say that $f$ is irreducible.
The question is, what are the irreducible maps $f:\lbrace 0,1 \rbrace^n \to \lbrace 0,1 \rbrace$ satisfying (1) and (2) ?
When $n$ is odd, there is an obvious solution of "choosing the candidate with the most votes". I couldn't find any other examples. 
My thoughts : 
This is very closely related to Arrow's theorem and similar results, of course. But everything I've found so far only dealt with voting systems with at least three outcomes.
Update: The question can also be expressed in "measure-theoretic" terms. Indeed, if for $A\subsetneq \lbrace 1,2,\ldots, n \rbrace$ we denote by $i_A$ the indicator uple of $A$ (i.e. the uple $u=(u_1,\ldots,u_n)$ with $u_a=1$ when $a\in A$ and $0$ otherwise) and put $\mu(A)=f(i_A)$, then for the "measure" $\mu$, (1) means that $\mu(A)\leq\mu(B)$ when $A\subseteq B$, and (2) means that $\mu(A^{c})=1-\mu(A)$ where $A^{c}$ denotes the complement of $A$.
 A: There are definitely other irreducible functions that satisfy the monotonicity and symmetry conditions; here is one example. (The numbers $F_k$ below are the Fibonacci numbers.)
Take $n=7$, and let voter $k$ have $F_k$ votes for $k=1,\ldots,7$; the winning candidate is the one who gets the larger number of votes. (This is well-defined, since $\sum_{k=1}^7F_k=33$ is odd.) This is clearly not straight majority rule, since $f(\langle 0,0,0,0,0,1,1\rangle)=1$, and it clearly satisfies the monotonicity and symmetry conditions. 
Suppose that $I\subseteq[7]$ is a coalition; clearly we must have $\sum_{k\in I}F_k\ge 17$, so at least one of $6$ and $7$ must belong to $I$. If $7\notin I$, then $5\in I$. Then for any $a=\langle a_1,\ldots,a_7\rangle\in\{0,1\}^7$ such that $a_5\ne a_6$, $f(a)$ depends on $a_7$, so we must have $7\in I$. 
Suppose that $6\notin I$, and that $a_k=1-a_7$ for $k\in I\setminus\{7\}$. If $\sum_{k\in I\setminus\{7\}}F_k\ge 9$, then voter $6$ has enough votes to determine the outcome, so If $\sum_{k\in I\setminus\{7\}}F_k\le 8$. But then $\sum_{k\in[7]\setminus I}F_k\ge 12$, and the voters not in $I$ have enough votes to determine the outcome. Thus, $6\in I$.
Suppose that $5\notin I$, and that $a_6\ne a_7$. If $\sum_{k\in[7]\setminus I}F_k\ge 9$, then the voters in $[7]\setminus I$ have enough votes to determine the outcome. Otherwise, $\sum_{k\in[7]\setminus I}F_k\le 8$, so 
$$12\le\sum_{k\in I\setminus\{7\}}F_k\le 15\;,$$
and voter $5$ has enough votes to determine the outcome when voter $7$ disagrees with the rest of $I$. Thus, $5\in I$.
It’s straightforward to check in similar fashion that $4\in I$, then that $3\in I$, and finally that $I=[7]$ and hence that $f$ is irreducible.
I’ve not yet really tried to see how much this idea can be generalized — quite a bit, I suspect.
