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? $\exists x\forall y ((x + y \in X) \wedge (x - y)\in Z)$
My solution:
(x+y) =X (x-y) = Z
x = 2 and y = 1
2+1 = 3 2-1 =1
x = 2 and y = 2
2+2 = 4 2-2 = 0
trying different x for y = 2
x = 1 and y = 2
1+2 = 3 1-2 = -2
Am I doing this right? The statement seems false to me.
Supposing that such $x$ exists, since it must work for every $y$ (in $U$ I suppose), it must work for $y=0$. So $x\in X$ and $x\in Z$. So it must be 1 or 3 but neither 1 nor 3 fits the property.