Multiset equivalence Let $a,b,c,d,e,f,g,h$ be natural numbers such that the multisets $\{a,b,c,d,a+b,c+d\}$ and $\{e,f,g,h,e+f,g+h\}$ are the same. Can we say that $\{a,b\}=\{e,f\}$ or $\{g,h\}$ ? and similarly $\{c,d\}=\{e,f\}$ or $\{g,h\}$ ? I have tried with case by case. Looks like it is correct. Is there an elegant proof for this ?
 A: Notice that since all elements are non-negative, either $a+b$ or $c+d$ will attain the maximal value of the multiset, similarly for $e+f$ or $g+h$. After possible relabeling/swaping of $(a,b)$ with $(c,d)$ and $(e,f)$ with $(g,h)$, we can assume a maximal element is attained by $a+b$ and $e+f$, and so
$$a+b=e+f\tag{1}.$$
Also, since the two multisets are equal, sums of their elements are also equal, so $2a+2b+2c+2d=2e+2f+2g+2h$, which means
$$a+b+c+d=e+f+g+h\tag{2}.$$
Then subtracting $(1)$ from $(2)$, we have
$$c+d=g+h\tag{3}.$$
By removing pairs of equal elements given by $(1)$ and $(3)$ from both multisets, we get
$$\{a,b,c,d\}=\{e,f,g,h\}.\tag{4}$$
If $a\in\{e,f\}$, then $(1)$ implies $b\in\{e,f\}$ (and vice-versa), then $\{a,b\}=\{e,f\}$. Then by $(4)$ the only other option is $a,b \in \{g,h\}$, but then $\{a,b\}=\{g,h\}$ (the other inclusion is given by the same reasoning starting with $g,h$, leaving us with $g,h \in \{a,b\}$).
Finally, since either $\{a,b\}=\{e,f\}$ or $\{a,b\}=\{g,h\}$, removing these from $(4)$ also gives $\{c,d\}=\{g,h\}$ or $\{c,d\}=\{e,f\}$, as desired.
Remark. The non-negativity is crucial, otherwise we'd have for example $(a,b,c,d)=(1,-1,1,-1)$,$(e,f,g,h)=(0,1,0,-1)$ as a counterexample. No other assumption was used, so the claim holds for all non-negative real numbers.
