We have the diamond lemma as follows:
Let $\rightarrow$ be relation on a set $P$. Let $\twoheadrightarrow$ be the reflexive transitive closure of $\rightarrow$ and $\sim$ the equivalence relation generated by $\rightarrow$ then if we have the following two conditions:
1) There is no infinite sequence $a_0\rightarrow a_1\rightarrow \ldots $
2) Let $a, b, c \in P$ be such that $a \rightarrow b$ and $a \rightarrow c$. Then there exists $d \in P$ such that $b \twoheadrightarrow d$ and $c \twoheadrightarrow d$.
Then every $\sim$-clas contains a unique element $a$ such that $a\not\rightarrow b$ for all $b\in P$.
Now I am told to proceed in the following way:
Consider $Q(a):=\{b\in P|a\twoheadrightarrow b\}$. Then let $R$ be the set of elements $a\in A$ such that $Q(a)$ doesn't have a least element.
I really have no idea where to go from here.
Thanks for any help
Let $Q(a)$ and $R$ be as in the OP. It will suffice to show the following lemma (which immediately allows one to define inductively a sequence contradicting property (1)) :
Lemma. If $r\in R$, then there is a $s\in P$ such that $r\to s$ and $s\in R$.
Proof of lemma. Suppose by contradiction that $I=\lbrace b\in P | r \to b\rbrace$ does not contain any element of $R$. Then for any $x\in I$, $Q(x)$ has a least element $\lambda(x)$. By property (2), for any $x,y\in I$ there is a $d_{x,y}\in Q(x)\cap Q(y)$. We then have $ d_{x,y}\twoheadrightarrow \lambda(x)$ and $ d_{x,y}\twoheadrightarrow \lambda(y)$, so applying (2) again there is a $\mu\in Q(\lambda(x))\cap Q(\lambda(y))$. But since $\lambda(x)$ and $\lambda(y)$ are least elements, we must have $\mu=\lambda(x)=\lambda(y)$ for any $x,y\in I$. It is now clear that $\mu$ is the least element of $Q(r)$, contradicting $r\in R$. The lemma is proved.