Equivalence Relations on Set of Ordered Pairs Let $\mathbb{R}$ be the relation on $\mathbb{Z} \times \mathbb{Z}$, that is elements of this relation are pairs of pairs of integers, such that $((a,b),(c,d)) \in \mathbb{R}$ if and only if $a-d = c-b$. Can anyone give me a start on how to solve it to be transitive, reflexive and symmetric?
Reflexive: $\forall (a,b):\Bigl[(a,b)\in \mathbb{Z\times Z} \to \bigl((a,b),(a,b)\bigr)\in R\Bigr]$
Symmetric: $\forall (a,b,c,d): \Bigl[\bigl((a,b),(c,d)\bigr)\in R \leftrightarrow \bigl((c,d),(a,b)\bigr)\in R\Bigr]$
Transitive: 
$\forall (a,b,c,d)\exists (e,f): \Bigl[\bigl((a,b),(e,f)\bigr)\in R\land \bigl((e,f),(c,d)\bigr)\in R \leftrightarrow \bigl((a,b),(c,d)\bigr)\in R\Bigr]$


Show that these properties hold (or not) when $\Bigl[\bigl((a,b),(c,d)\bigr)\in R \Bigr]\iff\Bigl[ a-d=c-b\Bigr]$



Hint: $[a-d=c-b] \iff [a+b=c+d]$
Reflexive: $((a,b),(a,b))\in R$ since $a-b=a-b$.

Symmetric:  Suppose that $((a,b),(c,d))\in R$ since $a-d=c-b$, hence  $c-b=a-d$. So  $((c,d),(a,b))\in R$.

Transitive:  Suppose that $((a,b),(c,d))\in R$ and $((c,d),(e,f))\in R$. Then  $a-d=c-b$ and  $c-f=e-d$, and hence $a-d+c-f=c-b+e-d$ which implies that $a-f=e-b$. So we have that $((a,b),(e,f))\in R$