If $R$ is reduced residue system then $\exists! b\in R$ such that $a\equiv b\pmod n$ 
If $R$ is a reduced residue system modulo $n$ and $\gcd(a,n)=1$ then $\exists! b\in R$ such that $a\equiv b\pmod n$.

Recall:
$$R=\{a\in C \mid \gcd(a,n)=1\}$$ Where $C$ Is a complete residue system modulo $n$
I didn't quite understand the theorem even when i‘ve flipped the paper so i can see the proof of it, let’s take an example to clarify things.
Example:
Let $C=\{0,1,2,3\}\implies R=\{1,3\}$ but where is $a$ here? If $a=3,n=4$ Then $b$ here is just $3$, and it’s obvious that $3\equiv 3\pmod 4$
, so my question is what is $a$.
 A: I believe the proper formulation is the below:

If $R$ is a reduced residue system modulo $n$ and $a\in\Bbb Z$ such that $\gcd(a,n)=1$, then $\exists b\in R$ such that $a\equiv b\mod n$.

Since we have already used up the variable name $a$ in the above statement, we will write $R=\{x\in C:\gcd(x,n)=1\}$.
The proof of this statement is simple: let $a\equiv y\in C$. Then $\exists k\in \Bbb Z|a=kn+y$. Thus $1=\gcd(a,n)=\gcd(kn+y,n)=\gcd(y,n)$ which implies $y\in R$.

Edit: If $\exists!b$ means "one and only one $b$", we should add that if $a\equiv b_1\equiv b_2\in R$, then $b_1=b_2$ as no two distinct elements of $C$ are congruent.
A: well, as $R \subset C$ and, by definition of complete residue system, then for any $a\in \mathbb Z$ there is a unique $b \in C$ so that $a\equiv b \pmod n$.
So this statement is equivalent to proving, if $a \equiv b \pmod n$ then $\gcd(a,n) = 1 \iff \gcd(b,n) = 1$
And that follows fairly straightforwardly.
$\gcd(a,n) = \gcd(a+kn, n)$ for all integers $k$.
And $a \equiv b \pmod n$ is defined as $n|b-a$ which means that there is an integer $k$ so that $a = b+kn$.   So $\gcd(b,n) = \gcd(a + kn, n) =\gcd(a,n)$.
....
Perhaps a stronger and more useful statement would be:

Lemma:  If $a\equiv b \pmod n$ then $\gcd(a,n) =\gcd(b,n)$.

