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Here is a counter-example: let $A=(1,0,0,0,0,1)$, $B=(1,1,1,0,0,0)$, $C=(0,0,1,1,0,0)$ and $D=(0,0,0,1,1,1)$. Take $Y=\{A,B,C,D\}$. Then:

• $PE=\{A,B,C,D\}$, so that $conv(PE)=conv(Z)$.

• The vector $(.5,0,.5,.5,0,.5)$ is in $conv(PE)=conv(Z)$, but it is not in $E$ since $(.5,.5,.5,.5,.5,.5)$ is also in $Z$.

(The second or the fourth coordinate can be deleted in the example, to get an example with only five coordinates. I am not sure whether that is smallest possible).

Here is a counter-example: let $A=(1,0,0,0,0,1)$, $B=(1,1,1,0,0,0)$, $C=(0,0,1,1,0,0)$ and $D=(0,0,0,1,1,1)$. Take $Y=\{A,B,C,D\}$. Then:

• $PE=\{A,B,C,D\}$, so that $conv(PE)=Z$.

• The vector $(.5,0,.5,.5,0,.5)$ is in $conv(PE)=Z$, but it is not in $E$ since $(.5,.5,.5,.5,.5,.5)$ is also in $Z$.

(The second or the fourth coordinate can be deleted in the example, to get an example with only five coordinates. I am not sure whether that is smallest possible).

Here is a counter-example: let $A=(1,0,0,0,0,1)$, $B=(1,1,1,0,0,0)$, $C=(0,0,1,1,0,0)$ and $D=(0,0,0,1,1,1)$. Take $Y=\{A,B,C,D\}$. Then:

• $PE=\{A,B,C,D\}$, so that $conv(PE)=conv(Z)$.

• The vector $(.5,0,.5,.5,0,.5)$ is in $conv(PE)=conv(Z)$, but it is not in $E$ since $(.5,.5,.5,.5,.5,.5)$ is also in $Z$.

Here is a counter-example: let $A=(1,0,0,0,0,1)$, $B=(1,1,1,0,0,0)$, $C=(0,0,1,1,0,0)$ and $D=(0,0,0,1,1,1)$. Take $Y=\{A,B,C,D\}$. Then:

• $PE=\{A,B,C,D\}$, so that $conv(PE)=conv(Z)$.

• The vector $(.5,0,.5,.5,0,.5)$ is in $conv(PE)=conv(Z)$, but it is not in $E$ since $(.5,.5,.5,.5,.5,.5)$ is also in $Z$.

(The second or the fourth coordinate can be deleted in the example, to get an example with only five coordinates. I am not sure whether that is smallest possible).

Here is a counter-example: let $A=(3,0)$, $B=(1,1)$ and $C=(0,3)$. Take $Y=\{A,B,C\}$. Then:

• $PE=\{A,B,C\}$, so its convex hull is a triangle.

• $Z$ is the same triangle.

• $E$ is the edge $AB$ of that triangle.

Update: oupps, this is not a counter-example to the question, since in the question coordinates are 0/1

Here is a counter-example: let $A=(1,0,0,0,0,1)$, $B=(1,1,1,0,0,0)$, $C=(0,0,1,1,0,0)$ and $D=(0,0,0,1,1,1)$. Take $Y=\{A,B,C,D\}$. Then:

• $PE=\{A,B,C,D\}$, so that $conv(PE)=conv(Z)$.

• The vector $(.5,0,.5,.5,0,.5)$ is in $conv(PE)=conv(Z)$, but it is not in $E$ since $(.5,.5,.5,.5,.5,.5)$ is also in $Z$.