$P$ be a poset with more than min$\{rs-r,rs-s\}$ elements where $r,s\in\Bbb{N}$. Prove that $P$ has an anti-chain of size $r$ or a chain of size $s$ Let, $(P,\le)$ be the poset.
I have begun to solve this in the following way-
Note that, $rs-r\le rs-s\iff r\ge s$
So, without loss of generality assume that $r\ge s$, then $\operatorname{min}(rs-r,rs-s)=r(s-1)$
As per the question $P$ has elements $\ge r(s-1)$.
So, let number of elements of $P$ is $r(s-1)+n$ where $n\in\Bbb{N}$.
Let us assume on contrary, $P$ neither has an anti-chain of size $r$ nor a chain of size $s$
i.e. for any $A$ of $P$ with $r$ elements, $\exists a,b\in A$ such that either $a\le b$ or $b\le a$.
And for any $C$ of $P$ with $s$ elements, $\exists x,y\in A$ such that neither $x\le y$ nor $y\le x$.
Now, I cannot use the number of elements of $P$ to get a contradiction from the above assumption.
Can anybody help me with this? Thanks for assistance in advance.
 A: Assume that $P$ has neither an antichain of size $r$ nor a chain of size $s$.
Let $\{ a_1, a_2, \dots, a_k \}$ be a maximal antichain in $P$.
Define $C_1$ to be a maximal chain in $P$ containing $a_1$, and for each $1 \le i < k$, let $C_{i+1}$ be a maximal antichain in $P \setminus (C_1 \cup \cdots \cup C_i)$ containing $a_{i+1}$.
You can verify that $k < r$, that $|C_i| < s$ for each $i$, that $C_1 \cup \cdots \cup C_k = P$, and that the $C_i$ are all disjoint. It then follows that
$$|P| = \sum_{i=1}^k |C_i| \le \sum_{i=1}^k (s-1) = k(s-1) \le (r-1)(s-1)$$
But
$$(r-1)(s-1) = (rs-r)+(1-s) = (rs-s)+(1-r)$$
so $|P| \le \mathrm{min} \{ rs-r, rs-s \}$, as required.
A: HINT: Use Dilworth’s decomposition theorem. I’ve finished the argument in the spoiler-protected block below.

 Let $A$ be an antichain in $P$ of maximum size; say $|A|=m<r$. By Dilworth’s theorem $P$ has a decomposition into $m$ disjoint chains, all of which are of length at most $s-1$, so $|P|\le m(s-1)<r(m-1)$, contradicting the choice of $P$.

