Is this true? $\forall x\exists y ((x + y \in X) \vee (x - y)\in Z)$ U = $\begin{Bmatrix}
0,1,2,3,4,5,6,7,8,9
\end{Bmatrix}$
X = $\begin{Bmatrix}
0,1,2,3,4
\end{Bmatrix}$
Z = $\begin{Bmatrix}
1,3,5,7,9
\end{Bmatrix}$
Is this true?
$\forall x\exists y ((x + y \in X) \vee  (x - y)\in Z)$
My solution:  (Before this i tried with y = 1 and y = 2)
y=3
x = 1 y = 3 --> 1+3 = 4 --> (X)
x = 2 y = 3 --> 3-2 = 1 --> (Z)
x = 3 y = 3 --> 3-3 = 0 --> (Z)
x = 4 y = 3 --> 4-3 = 1 --> (Z)
x = 5 y = 3 --> 5+3=8 or 5-3 = 2 
So  y = 3 is not true because 8 is not in (X) and 2 is not in (Z).
Do i need to test this for every  other y = 4,5,6...9 to verify if this is true or not? Or is there a better faster way to do this?
 A: First, the statement you are considering is $\forall x \in U \left( \exists y \in U \left( \left( x + y \right) \in X \vee \left( x - y \right) \in Z \right) \right)$.
Observe that if $x \in X$ or $x \in Z$, then $y = 0 \in U$ works. For the other numbers, namely $6$ and $8$ observe that $6 - 5 = 1 \in Z$ and $8 - 7 = 1 \in Z$. Hence, the statement is true.
First, you need to "choose" an $x \in U$ and then corresponding to this you need to "find" a $y \in U$ so that either the sum is in $X$ or the difference is in $Z$. What you were trying to do in your attempt was that for every pair of elements $x$ and $y$ in $U$, you were trying to figure out if the sum was in $X$ or the difference was in $Z$. This is not to be done for the statement we have.
A: The statement says, in plain english:

For every $x$ there exists a $y$ such that either $x+y$ is in $X$ or $x-y$ is in $Z$.

Therefore, you must prove the statement for all values of $x$, but the $y$ value doesn't need to be the same. 
For $x=1,2,3,4$, you can select $y=3$.
For $x=5$, the value $y=3$ is not good, but that doesn't mean there doesn't exist some other value of $y$ that is good. In fact, if $x=5$, we can take $y=4$ to prove that the statement holds for $x=5$.
