$x\sim y$ implies $x+a\sim y+a$ for any $a\geq0$ and $x,y\in\mathbb R^n$, then the preference is linear? $x,y,a$ are vectors in $\mathbb R^n$

We say $a\geq0$ if all directions of the vector $a$ is greater or equal to zero.

We want to prove (or disprove by counterexample) that: 

Suppose $x\sim y$ implies $x+a\sim y+a$ for any $a\geq0$ and $x,y\in\mathbb R^n$, 

Then the preference is linear. 



One definition of linear preference is that $x\sim y$ implies $x+a\sim y+a$ for any $x,y,a$.
It is not true. Let us consider $\mathbb{R}^2$ so bundles are $x = (x_1,x_2)$.

Consider the preference:

(i) If $x_1 \leq 0$, preferences are lexicographic, i.e. 
$$
x \succ y \Leftrightarrow 
\begin{cases}
x_1 &gt; y_1 \\
\text{ or } \\
x_1 = y_1 \text{ and } x_2 &gt; y_2
\end{cases}
$$ 
(ii) If $x_1 \geq 0$, $u(x_1,x_2)=x_1+x_2$.

Notice that no indifference occurs on $\mathbb{R}_{&lt;0} \times \mathbb{R}$, and the required condition holds on $\mathbb{R}_{\geq 0}\times \mathbb{R}$ since it is linear there.

However, $(0,4) \sim (2,2)$, choose $a = (-1,-1)$, we get 
$$(2-a,2-a) = (1,1) \sim (0,2) \succ (-1,3) = (0-a,4-a)
$$
hence the second condition fails.