first of all I'm sorry for my bad English and second I'm sorry for my mistakes of understanding the following topic, I still going to school and learning this for interest.
The topic is Myhill-Nerode and the equivalence classes of a regular or non regular language.
I know that every element of a equivalence class by Myhill-Nerode fulfills this property:
$ \equiv_{A} \triangleq\{(x, y) | \forall z \in \Sigma^{*} \cdot(x z \in A \leftrightarrow y z \in A)\} $
If I understand this right, than a equivalence class consist of element (words) $x$ which we can expand with a word $y$ but for all words $x$ and $y$ of the same class must apply, adding a word $z$ to them both must be in or out of the language.
Hope that is right until now.
Now I will show you my problem:
I have the language (its from a book):
$ \mathrm{B} \triangleq\left\{73 \mathrm{a}^{n} 7 \mathrm{b}^{\mathrm{m}} | \mathrm{n}, \mathrm{m} \in \mathbb{N} \wedge \mathrm{n}=\mathrm{m}+2\right\} $ with $ \Sigma_{\mathrm{M}} \triangleq\{\mathrm{a}, \mathrm{b}, 3,7\} $
And a complete solution:
$1: [\lambda]\equiv_{B}=\{\lambda\} $
$2: [7]\equiv_{B}=\{7\} $
$3: \left[73 a^{k}\right] \equiv_{B}=\left\{73 a^{k}\right\} $ für $ k \in \mathbb{N} $
$4: \left[73 a^{l+2} 7\right] \equiv_{B} =\left\{73 \mathrm{a}^{\imath+2+n} 7 \mathrm{b}^{\mathrm{n}} | \mathrm{n} \in \mathbb{N}\right\} \quad $ für $ l \in \mathbb{N} $
$5: [3]_{\equiv \mathrm{B}}=\Sigma^{*} \backslash\left([\lambda]_{\equiv \mathrm{B}} \cup[7]_{\equiv \mathrm{B}}\right. \cup\left(\bigcup_{k \in \mathbb{N}}\left[73 a^{k}\right] \equiv_{B}\right) \cup \left(\bigcup_{\mathfrak{l} \in \mathbb{N}}\left[73 \mathrm{a}^{\ l+ 2} 7\right] \equiv_{\mathrm{B}}\right) ) $
So in $1$ they build a class of the empty word $\lambda$ and $z = B$ has to be the language by her self to be in the language?
In $2$ they build they build the class of $7$ and z has to be something like this $ z = \left\{3 \mathrm{a}^{n} 7 \mathrm{b}^{\mathrm{m}} | \mathrm{n}, \mathrm{m} \in \mathbb{N} \wedge \mathrm{n}=\mathrm{m}+2\right\}$ to be in the language.
In $3$ they build a class or better infinitely many classes. But here is my problem. I cannot find a $z$ which is working for all classes.
For example we have the words $x_i$ and $z_i$
$x_1 = 73 \to z_1= {a}^{n+2}7b^n$ with $n\in N$
$x_2 = 73a \to z_2= {a}^{n+1}7b^n$ with $n\in N$
$x_3 = 73a^2 \to z_3= {a}^{n}7b^n$ with $n\in N$
$x_4 = 73a^3 \to z_4= {a}^{n}7b^n+1$ with $n\in N$
and so on.
But why is this ok? I Mean they are 2 words in this class for example $x_1$ and $x_2$ who $x_2$ would not be in $B$ with $z_1$.
I hope you can tell me on a simple and understanding way how those classes by Myhill work and how i can find them without making big mistakes.