**Definition:**
Let $G$ be a graph. A subset $C \subseteq V(G)$ is a vertex cover of
$G$ if for each $e \in E(G)$, $e\cap C \neq \phi$. If $C$ is minimal
with respect to inclusion, then $C$ is called minimal vertex
cover of $G$.


Let $G$ be a graph with $V(G)=\{z\_1,\ldots,z\_n\}$. 
Let $H$ be the new graph $H$ on new vertices $\{z\_{1,1},z\_{1,2},\ldots ,z\_{n,1},z\_{n,2}\} $ and edge set of 
$H$ be 
$E(G)=\Big\{ \{z\_{a,b},z\_{c,d}\} \mid \{z\_{a},z\_c\} \in E(G) \text{ and } b+d \leq 3\Big\}$.


One can see that if $M\_1=\{z\_{i\_1},\ldots,z\_{i\_r}\}$ is a minimal vertex cover of $G$ , then $M\_2=\{z\_{i\_1,1},z\_{i\_1,2},\ldots,z\_{i\_r,1},z\_{i\_r,2}\}$ is a minimal vertex cover of $H$.



> 
> Is $M\_2$ the largest cardinality of minimal vertex cover of $H$?
> 
> 
> 



Minimal vertex cover
Not necessarily. E.g., the set $\{z\_{i,1}\colon 1\leq i\leq n\}$ is also a minimal vertex cover of $H$ (whenever $G$ has no isolated vertices), and it may have more elements (e.g., if $G$ is a star, and $M\_1$ consists of its center).


In fact, all minimal covers of $H$ can be pbtained in the following way. Notice that, if $C\_2$ is a cover of $H$, then $C\_1=\{z\_i\colon z\_{i,1}\in C\_2\}$ is necessarily a cover of $G$. So, let us start with a (not necesarily minimal) cover $C\_1$ of $G$, and include into a sought cover $C\_2$ of $H$ all vertices $z\_{i,1}$ with $z\_i\in C\_1$. We need also to include all $z\_{j,2}$ which are neighbours of some $z\_{k,1}$ with $z\_k\notin C\_1$. 


This constitutes a cover of $H$, but not necessarily a minimal one. Say that a vertex in a covering is *necessary* if it cannot be removed. Then all vertices $z\_{j,2}\in C\_2$ are necessary, as well as all vertices corresponding to necessary vertices of $C\_1$. On the other hand, a vertex of $C\_2$corresponding to unnecessary vertex in $C\_1$ is necessary if it has a neighbor $z\_{j,2}$ which is not in $C\_2$; that neighbor coresponds to another unnecessary neighbor in $C\_1$.


Thus, the condition that $C\_2$ is monomal is; *Each unnecessary vertex in $C\_1$ has an unnecessary neighbor*.


**[EDIT OF ADDENDUM]** Sorry, the previous argument was wrong. It showed only that the conjecture from the comment holds whenever each induced subgraph of $G$ has a minimal vertex cover containing at least half of its vertices. However, there exist graphs where each minimal vertex cover has less than a half of vertices, and each such graph is a counterexample to what you wanted.


E.g., take $K\_3$, and attach to each its vertex two degree 1 vertices (so we have 9 vertices at all). Each its minimal vertex cover contains two vertices of $K\_3$, hence it contains either the third vertex as well, or two its leaves. Hence $M\_1$ may contain at most 4 vertices. On the other hand, the graph $H$ admits a minimal vertex cover of 9 vertices, as described above.


