Using the fact that \begin{align*}(a|b \ \land \ a|c) &\implies a|(b+c) \\ a|b &\implies a| bc \\ (a|b \ \land \ b|c) &\implies a|c \end{align*}
then I want to prove $\equiv_d$ is an equivalence relation on $\mathbb Z$ for every positive integer $d$
I'm pretty confused on this, but so far what I have done is attempt to show reflexivity and symmetry, but not yet transitivity and I don't know how to apply the predicates above. For reflexivity I said $dRd \implies d-d \in \mathbb Z, d\in \mathbb Z^+$, and since $0\in \mathbb Z$ then there is a reflexive relation. For symmetry I said if we take any $k\in \mathbb Z$, then $dRk \implies kRd$. So then $d-k \in \mathbb Z, k-d \in \mathbb Z$. Since $d-k=c, k-d=-c$ then symmetry is satisfied. And it is this point I don't know how to continue further.
Reflexive: For every $a \in \mathbb Z$, $a\equiv a\pmod d$.
Pf: $d= 1\cdot d$ so $d|d$. So $d|d\times 0 = 0=a -a$ for any $a\in \mathbb Z$. So $a\equiv a \pmod d$.
Definition of $a\equiv a \pmod d$ is that $d|(a-a)$ or in other words $d|0$. Does $d|0$?
Well $d|d$ so $d|d\cdot 0=0=a-a$. So yes it does. So $d|a-a$ so $a \equiv a \pmod d$ for all $a\in \mathbb Z$. So $\equiv_d$ is reflexive.
Symmetric: For every $a,b \in \mathbb Z$ where $a\equiv b\pmod d$ then $b \equiv a \pmod d$.
Proof: $a\equiv b\pmod p$ means $d|a-b$. So $d|-1(a-b) = b-a$. So that means $b\equiv \pmod d$. SO if $a \equiv b \pmod d$ then $b\equiv a \pmod d$. So $\equiv_d$ is symmetric.
Transitive. If $a \equiv b\pmod d$ and $b\equiv c \pmod d$ then $a\equiv c\pmod d$.
Proof. If $a \equiv b \pmod d$ then $d|(a-b)$. And if $b\equiv c\pmod d$ then $d|(b-c)$.
So $d|(a-b) + (b-c) = a-c$.
So $a\equiv c \pmod d$. So $\equiv_d$ is transitive.