Myhill-Nerode equivalence classes of $\{1^n0^n\}$ I have the following task and its solution.
Question

Given the language
$$ A \triangleq\left\{1^{n} 0^{n} \mid n \in \mathbb{N}\right\} \text { with } \Sigma_{A} \triangleq\{1,0\}, $$
give all equivalence classes of the Myhill-Nerode relation.

Solution


$[1^k]_{\equiv A} = \{1^k\}$ for $k \in \mathbb N$.
$[1^\ell0]_{\equiv A} = \{1^{\ell+i-1} 0^i \mid i \geq 1\}$  for $\ell \in \mathbb N^+$.
$[0]_{\equiv A} = \{ 0x, 1^n 0^m, x01 y \mid x,y \in \Sigma_A^* \land n,m \in \mathbb N^+ \land m &gt; n \}$.


What are they doing in the second bullet of the solution?
The first and third bullets are clear. In the first they construct the
equivalence classes of all $1$'s including $\lambda$.
In the third bullet they construct just one equivalence class with all the words which are not in the language $A$.
But what are they doing in the second bullet?
What does the exponent $\ell+i-1$ mean?
Why don't they write $[1^n0^m]_{\equiv A} = \{1^n 0^m \mid n,m \in \mathbb N^+\}$?
I think it makes sense to answer your subquestions out of order.




  Why they don't write $[1^n0^m] = \{1^n 0^m | n,m \in N^+\}$


They're supposed to be equivalence classes. $[1^n0^m]$ would break up lots of classes, and duplicate lots of the states from line 3.




  What does the Exponent $n+i-1$ mean?


Firstly, $n+i-1 \ge i$, so there are at least as many $1$s as $0$s.

Secondly, $n$ is the parameter which defines the class.

So $\left[1^{\mathrm{l}}0\right]_{\equiv \mathrm{A}}$ is the set of states which need $0^{l-1}$ to reach the accepting state.