Prove that if $x\leq y$ then $x+z\leq y+z$ Does the below proof looks correct for the above question?

$$\text{Either x = y or (y - x) $\in $ }\mathbb{R}^+$$
$$\text{Case 1: x=y $\forall $z$\in $}\mathbb{R}^+$$
$$\text{x+z=y+z}$$ 
$$\text{Case 2: x $&lt;$ y $\forall $z$\in $}\mathbb{R}^+$$
$$\text{(y + z) - (x + z) = y + z - x - z = (y - x)$\in $}\mathbb{R}^+$$
And so $(y-x)$ is a positive number implies y >0

In particular we have $$x\leq y $$

Could someone verify?
I understand and agree with every step of your proof.  But I think the way you presented your proof as a list of steps with few words makes it difficult to read.  I think you should write out your proof like a story.  It's easier to follow than having step after step in a list.
Here's how I would have written the proof using the same steps as you, but presented in a more readable way:

Problem: Let $x, y, z \in \Bbb R$ and suppose $x \leq y$.  Prove $x + z \leq y + z$.
Proof: Since $x \leq y$, then either $x = y$, or $x &lt; y$.
If $x = y$, then for any $z$, $x + z = y + z$, so it follows that $x + z \leq y + z$.
Now, if $x &lt; y$, then $y - x &gt; 0$.  This implies $y + 0 - x &gt; 0$ which implies $y + (z - z) - x &gt; 0$, and this implies $(y + z) - (x + z) &gt; 0$.  Finally, this implies $y + z &gt; x + z$, so we get that $y + z \geq x + z$.
Thus, we have shown in every case that $x + z \leq y + z$, as desired.
I was not sure how to give a definite answer to this so I decided to show a proof that I know works. Let us say that y-x=a, (x+z)-(y+z)=b, and a is greater than or equal to 0. This means that if our proof is correct, b will be greater than or equal to 0. So, what happens with (x+z)-(y+z)? Well, it also equals a because that anything minus itself equals 0, and anything plus 0 equals itself. So, not only is b greater than or equal to 0, but also a=b! Sorry this dose not answer your question, but I hope it is useful anyway!