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How to show that Max-cut$_{dec}$ is NP-complete using Mon-NAE-3SAT ?

Mon-NAE-3SAT definition :

An instance is a m clauses of three positive literals (no complemented variable) $x \vee y \vee z$ The question: is whether there is a Boolean assignment that ensures that each clause contains at least one variable

Max-cut$_{dec}$ definition :

Let $G = (V,E)$ be an undirected graph. A cut (or cut) is a partition of $V$ into two subsets $V_1, V_2$. The size of a cut, note $c(V_1,V_2)$, is the number of edges $(u,v)$ of $E$ such that $u \in V_1$ and $v \in V_2$ or, conversely, $u \in V_2$ and $v \in V_1$.

The Max-cut optimization problem is finding the largest cut. In its decision version, Max-cut$_{dec}$, each instance also includes a $k$ size objective and you have to decide if it exists a cut of $k$ size or more.

actually I'm stuck on how to show :

  • that if $x, y, z$ are on the same side of the cut then the gadget contains at most 4 edges counted in the cut.
  • that if the three points $x,y,z$ are not all on the same side of the cut, then we can place points $a,b,c$ in $V_1$ or $V_2$ so that the gadget contains 5 edges counted in the cut.

I put the gadget below which represents an $x \vee y \vee z$ clause.

enter image description here

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