I would like to rigorously prove the following result for integers $\mathbf{Z}$. This is exercise 4.1.1, page 81, from Analysis-I by Terrence Tao.
Verify that the definition of equality of integers is both reflexive and symmetric.
I would like someone to verify my proof. I just want to make sure, that I am not making implicit assumptions or trivializing my proof. I also want to avoid any circularity.
Proof. (My attempt).
Let $a,b,c,d$ be natural numbers. We define integers $x = a - b$ and $y = c - d$.
The equality relation $=_{\mathbf{Z}}$ on the integers $\mathbf{Z}$ is defined to be the set of all ordered pairs $(x,y)$ given by
$R := \{ (x,y) | (a + d) = (b + c); \text{ such that } x = a - b, y = c - d\}$
(1) Reflexive. Clearly, $a + b = a + b$ for natural numbers $a,b$, so $a - b = a - b$. This implies $x = x$ for all integers $x$.
(2) Symmetric. Moreover, if $x = y$, then $a + d = b + c$, it implies $b + c = a + d$, so $y = x$.
Thank you so much,
Quasar.