Set theory question regarding $A\overline{\sim} B = \{x+y : \in A\times B\}$ Given $A,B \in P(N)$ We mark $\overline{\sim}$ as
$$A\overline{\sim} B = \{x+y : \langle x,y\rangle\in A\times B\}$$
Now, order R will be as following
$ARB$ iff $\exists M\in P(N)$ so $A\overline{\sim} M=B$.
The question is 
R is reflexive? symmetric? anti-symmetric? transitive?
I think R is partial order, i.e reflexive, anti-symmetric and transitive.
and I need to prove it.
I think my proof is correct for reflexive so let`s focus on anti-symmetric and transitive.
Just for make it sure - N is Set of Natural numbers, including zero.
Here what I done for proving transitive:
Let $A,B,C \in P(N)$ and assume ARB and BRC. 
ARB so exists $T \in P(N)$ so $A\overline{\sim} T=B$
BRC so exists $S \in P(N)$ so $B\overline{\sim} S=C$
Let define M as following set: $\{K+P | k\in T$ and $p\in S\}$
$S,T\in P(N)$ so $M\in P(N)$ we need to show $A\overline{\sim} M=C$
Let $q\in A\overline{\sim} M$ 
here I got stuck, since I have no idea how to show that $q\in C$.
Might this R is not transitive?
Anti-Symmetric:
Let $A,B\in P(N)$ and assume ARB and BRA we need to show A=B.
let $a\in A$
ARB so exists $M\in P(N)$ so $A\overline{\sim} M=B$
and here I got stuck.
Can we say that M is a set with just a zero element? 
we have exists as not as target so we can`t choose it, I think..
Might R is not anti-symmetric?
Any help would be appreciated.

The usual notation for $\overline{\sim}$ is +.
Show + is associative and communitive.   
Clearly, A + {0} = A, so ARA.
Assume ARB, BRC.  Thus exists K,L with A + K = B, B + L = C.
As A + (K + L) = (A + K) + L = B + L = C, ARC.  
