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.

You got off to a great start with your transitivity proof!
Note that $M=T\overline{\sim} S.$ Having chosen $q\in A\overline{\sim}M,$ we know that $q=a+m$ for some $a\in A$ and some $m\in M.$ Since $M=T\overline{\sim} S,$ then $m=t+s$ for some $t\in T$ and some $s\in S,$ whence $$q=a+(t+s)=(a+t)+s.$$ Since $a\in A$ and $t\in T,$ then $a+t\in A\overline{\sim} T=B,$ so since $s\in S,$ then $q\in B\overline{\sim} S=C.$ Thus, $A\overline{\sim}M\subseteq C.$
On the other hand, suppose that $q\in C=B\overline{\sim} S,$ so that $q=b+s$ for some $b\in B$ and $s\in S.$ Since $B=A\overline{\sim} T,$ then $b=a+t$ for some $a\in A$ and $t\in T,$ so that $$q=(a+t)+s=a+(t+s),$$ and since $M:=T\overline{\sim}S,$ then $q\in A\overline{\sim}M.$ Thus, $C\subseteq A\overline{\sim}M,$ so $C=A\overline{\sim}M,$ and so $ARM,$ as desired.

You started the proof of antisymmetry just as I would have, but you didn't follow through. Indeed, $A\overline{\sim}M=B$ for some $M,$ but also $B\overline{\sim}L=A$ for some $L.$ Thus, $a=b+l$ for some $b\in B$ and $l\in L.$ Furthermore, since $b\in B=A\overline{\sim} M,$ then $b=a'+m$ for some $a'\in A$ and $m\in M,$ so that $a=(a'+m)+l=a'+(m+l).$ It follows that $A\subseteq A\overline{\sim}(M\overline{\sim}L).$
On the other hand, take $x$ to be any element of $A\overline{\sim}(M\overline{\sim}L),$ so that $x=a+(m+l)=(a+m)+l$ for some $a\in A,m\in M,l\in L.$ Since $a+m\in A\overline{\sim}M=B,$ then $x\in B\overline{\sim}L=A.$ Thus, $A\overline{\sim}(M\overline{\sim}L)\subseteq A,$ and so $A=A\overline{\sim}(M\overline{\sim}L).$
Now, if $A=\emptyset,$ it's clear that $A\overline{\sim}M=\emptyset$ by the definition of $\overline{\sim},$ so $B=A$ in that case. Suppose that $A$ is not empty, and let $a_0$ be the least element of $A.$ Since $a_0\in A=A\overline{\sim}(M\overline{\sim}L),$ then $a_0=a+(m+l)$ for some $a\in A,m\in M,l\in L.$ Consequently, $a\le a_0,$ so that $a=a_0$ by our choice of $a_0$ as the least element of $A.$ Thus, since $m,l\in\Bbb N,$ we conclude that $m=l=0.$ From this, we know that $0\in M$ and $0\in L.$
Thus, taking any $a\in A,$ we have that $a=a+0\in A\sim M=B,$ so $A\subseteq B.$ Similarly, $B\subseteq A,$ and so $A=B,$ as desired.
