Interpreting functions for poset I'm stuck at a question with the following expression of poset $(\mathcal{N}, \sqsubseteq)$:
$$\forall x, y \in \mathcal{N}: x \sqsubseteq y \mbox{ iff } (x = y \vee x = \bot_{\mathcal{N}} \vee y = \top_{\mathcal{N}}).$$
$\mathcal{N} = \mathcal{N} \bigcup\{ \bot_{\mathcal{N}} , \top_{\mathcal{N}}  \} $ consists of all natural numbers with two elements $ \bot_{\mathcal{N}} $ and $ \top_{\mathcal{N}} $, and I'll need to find the poset height.
I don't know how to interpret this expression though.

This poset is the set of all natural numbers with equality relation, so any two distinct elements $x, y \in \mathcal{N}$ are incomparable, together with the least $\bot_\mathcal{N}$ and the greatest $\top_\mathcal{N}$ elements. To find its height you should find the largest cardinality chain, that is a subset in which any two elements are comparable. Can you take it from there?
Hint: Since any two distinct $x, y \in \mathcal{N}$ are incomparable, every chain in this poset can contain at most one element from $\mathcal{N}$. You can start by trying to find the chains of cardinality $1, 2$ and so on. It can be useful to draw a Hasse diagram for this poset. 
Updated. Here is some definitions (which you are assumed to be familiar with given that problem). Let $(P, \sqsubseteq)$ be a poset. Two elements $x, y \in P$ are comparable if either $x \sqsubseteq y$ or $y \sqsubseteq x$.
A chain $C$ is a subset of $P$ such that any two elements $x, y \in C$ are comparable. The height of $P$ is the largest cardinality (or size) of a chain in $P$.
So roughly, to find the height of your poset you should check all chains in it and find the chains of the largest size. Note the following:


*

*If $C$ is a chain, then $C \cup \{\top_\mathcal{N}\}$ and $C \cup \{\bot_\mathcal{N}\}$ are also chains. This is because the least and the greatest elements are comparable with any element of a poset.

*If $x, y \in \mathcal{N}$ and they are both not equal to $\top_\mathcal{N}$ and $\bot_\mathcal{N}$, then by definition of $\sqsubseteq$ we have $x \sqsubseteq y$ if and only if $x = y$.


From 1 it follows that if you have some chain, you can always add the least and the greatest elements to it obtaining new (and in general larger) chain. From 2 it follows that any chain in your poset can't contain two distinct natural numbers from $\mathcal{N}$ since they are not comparable. It means that the chain of maximal cardinality has at most $3$ element: $\top_\mathcal{N}$, $\bot_\mathcal{N}$ and some natural number $x \in \mathcal{N}$. And obviously there are such chains in your poset. So its height is $3$.
