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Let $\succeq$ be a preference relation on $X$. Is it true that $x \succeq y$ if and only if $\lnot (y \succ x)$?

I think it is true and my proof is as follows. To prove $\implies$ direction, we have $\lnot (y \succ x) \equiv \lnot(y \succeq x)$ or $x \succeq y$. So clearly if we assume $x \succeq y$, then $\lnot (y \succ x)$ is true. To prove $\impliedby$ direction, assume $\lnot (y \succ x)$ is true. Then either $\lnot(y \succeq x)$ or $x \succeq y$. If $x \succeq y$ then we are done. If $\lnot(y \succeq x)$, then since $\succeq$ is a preference relation it must be complete, which implies $x \succeq y$.

Is my proof correct? Also, in general, I just want to ask, is it true that whenever we have a preference relation $\succeq$, instead of proving all these facts, can we "informally" pretend $\succeq$ is like the inequality $\ge$, and everything works analougously? For example, we could prove that $\lnot (x \sim y)$ if and only if $x \succ y$ or $y \succ x$, but if we just wanted to use deduce this fact without actually proving it, we can clearly see it from $\lnot(x=y)$ if and only if $x>y$ or $y>x$.

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  • $\begingroup$ Welcome to Econ.SE. I have the feeling this is a duplicate. Those with more experience in these questions could be able to find one. I'm not very familiar with the topic. Maybe this one? $\endgroup$ – luchonacho Sep 4 '17 at 12:16
  • $\begingroup$ Some of your argument seems to be using the result desired to prove itself. "$\lnot (y \succ x) \equiv \lnot(y \succeq x)$ or $x \succeq y$" ? $\lnot (y \succ x) \implies \lnot (y \succ x)$ is what you are trying to show in the first place. $\endgroup$ – Kitsune Cavalry Sep 4 '17 at 13:40
  • $\begingroup$ I think your proof is correct. Your intuition in the last paragraph is largely correct as well. But be careful about using transitivity of $\succsim$ for sequences of $x$'s; you may need to invoke continuity for $\succsim$ to behave like $\ge$ on the reals. $\endgroup$ – Herr K. Sep 8 '17 at 4:07
  • $\begingroup$ @KitsuneCavalry: Regarding the point you raised, I think OP did get it correctly. Strictly preference $\succ$ is defined to be $y\succ x \equiv y\succsim x \text{ and } \lnot(x\succsim y)$; so the negation $\lnot(y\succ x)$ is exactly what OP wrote $\lnot(y\succsim x)\text{ or }x\succsim y$. $\endgroup$ – Herr K. Sep 8 '17 at 4:11
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If we define $\succeq$ on $X$ as the primitive, then the strict preference relation is derived as $$ x\succ y ~\Leftrightarrow~ x\succeq y ~\text{ but not }~ y\succeq x.$$

Note that you can view this as a definition of $\succ$. Now we prove $y\succeq x$ iff not $x\succ y$:

$(\Rightarrow)$ If $y\succeq x$, by definition, $x\succ y$ is not true;

$(\Leftarrow)$ If not $x\succ y$, then there are only three possibilities: (i) $ x\succeq y$ and $y\succeq x$; (ii) not $x\succeq y$ but $y\succeq x$; and (iii) neither $x\succeq y$ nor $y\succeq x$. Case (iii) can be ruled out if we assume completeness of $\succeq$. Therefore, in both (i) and (ii), we have $y\succeq x$.

It's worth mentioning that Kreps (1990) introduces another approach to define preference relations on $X$. He takes $\succ$ as the primitive and then derives the weak preference relation as: $$x\succeq y ~\Leftrightarrow~ \text{not }~ y\succ x.$$

In this case, there is no need to prove the equivalence since it is a definition.

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