There are two ways of deriving a preference relation from $\rhd$.
Lemma. For all $x,x\in X$,
$(x,x)\equiv(y,y)$.
Proof.If $(x,x)\rhd(y,y)$, then axiom 2
implies $(y,y)\rhd(x,x)$, contradicting asymmetry. ∎
Choose $a\in X$ and define $x\succ_1y$ iff $(x,y)\rhd(a,a)$. Define
$x\succ_2y$ iff $(x,a)\rhd(y,a)$. A good problem is to show that
both of these definitions are well-defined, that is, the choice of
$a$ does not matter, that these are preference relations, and that
$\succ_1{}={}\succ_2$. What other ways of defining preference relations
can you think of?
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