Sequence of integers such that the sum of two disjoint subsequences is different I would like to know if it is possible to generate a sequence of integers $S = x_0, ..., x_n$ such that, if $Q$ and $P$ are two disjoint subsets of $S$ (even of different cardinality), and $C = c_0, ..., c_{|P|}$ is a sequence of integers greater than 0, then
$$\sum_{x_i \in Q} x_i \neq \sum_{x_j \in P} c_j \cdot x_j$$ for every $Q$, $P$, and $C$.
Thanks :)
 A: Based on Leo's observations, I'm assuming we can't have $Q = \emptyset$ and all $c_i = 0$. Also I'm assuming you can't have $P = Q = \emptyset$ as this would also lead to a trivial equality in all cases.
Then unfortunately my answer is still disappointing and trivial, if I've understood the question properly, so maybe you can add more constraints to make the question more interesting. Define $S$ as:
$S = \{2, 3\}$
If $P$ is the empty set, then the inequality holds for all choices of $Q$ because, assuming $Q \neq \emptyset$, and since each $x_i > 0$, then the LHS is positive and the RHS is zero. If $P$ is the whole set, then $Q$ must be the empty set, due to disjointness. Then since there is some $c_i \neq 0$ and since each $p_i \neq 0$ then the RHS is at least $p_ic_i > 0$.
So we're left to show the singleton set cases where $P = \{p\}$ and $Q = \{Q\}$ with $p \neq q \in S$. In that case we just need to show this is impossible:
$q = cp$ with $q \neq p \in S$ and $c$ some arbitrary integer. Let's show it's impossible towards a contradiction. Clearly $c > 0$ because $q > 0$. But then we have a contradiction. Since $p, q$ are prime and $p, c > 0$, we have $q$, a prime number, equal to the product of two positive integers.
Note: If you were asking "Is this possible for all $n$" then the above only shows it's possible for $n = 2$, with of course the extra constraints we added to remove triviality.
