Generating a set of sums Let $x_1,\dots,x_n$ be (not necessarily distinct) real numbers, and let $A$ be a multiset formed by taking the $2^n-1$ sums of nonempty subsets of the numbers. For example if $(x_1,x_2)=(1,1)$ then $A=\{1,1,2\}$, and if $(x_1,x_2,x_3)=(1,2,-3)$ then $A=\{-3,-2,-1,0,1,2,3\}$.
Obviously a multiset $A$ of $2^n-1$ numbers need not be generated by some $n$ numbers in this way. Or it can be generated in more than one way, for example $(x_1,x_2,x_3)=(-1,-2,3)$ generates the same set as above. If we know that $0\not\in A$, is it still possible that $A$ is generated by more than one multiset of $n$ numbers?
 A: Suppose $A$ is generated by $x_1,x_2,\dots,x_n$ and also by $y_1,y_2,\dots,y_n$. Then for each subset $I$ of $\{\,1,2,\dots,n\,\}$ there's a corresponding subset $J$ of $\{\,1,2,\dots,n\,\}$ such that $$\sum_{i\in I}x_i=\sum_{j\in J}y_j$$ This is a (homogeneous) system of linear equations with rational coefficients, so if it has a (non-trivial) solution, then it has a rational (non-trivial) solution. 
Now if there's a rational solution, then, multiplying by a common denominator, there's an integral solution. Let's call it $a_1,a_2,\dots,a_n$ and $b_1,b_2,\dots,b_n$. Then we have $$(1+x^{a_1})(1+x^{a_2})\cdots(1+x^{a_n})=(1+x^{b_1})(1+x^{b_2})\cdots(1+x^{b_n})\tag1$$ We may assume no $a_i$ equals any $b_i$, as otherwise we could cancel one term on each side and get a solution with a smaller value of $n$. 
One or more of the $a_i$ might be negative. If so, then one or more of the $b_i$ must also be negative, and the sum, $-S$, of the negative $a_i$ must equal the sum of the negative $b_i$, since it will be the smallest element of $A$. Then we can multiply both sides of (1) by $x^S$ to get $$(1+x^{c_1})(1+x^{c_2})\cdots(1+x^{c_n})=(1+x^{d_1})(1+x^{d_2})\cdots(1+x^{d_n})$$ where every exponent is a positive integer. 
Now let $c$ be the largest of the $c_i$, $d$, the largest of the $d_i$. If $c<d$, then the right side vanishes at $x=e^{\pi i/d}$, and the left side doesn't, contradiction. Similarly if $c>d$, so we must have $c=d$. Now we can cancel $1+x^c$ on the left and $1+x^d$ on the right, and repeat the argument. It follows that the $c_i$ are just a permutation of the $d_i$. 
Now go back to (1). I claim that there must be a nonempty collection of the $a_i$ (hence, also of the $b_i$) summing to zero. But I'm having trouble seeing how to finish this off, so I'll leave it here until I can get back to it, or until someone else sees how to complete the argument. 
