<p>If we define $\succeq$ on $X$ as the primitive, then the strict preference relation is derived as
$$ x\succ y ~\Leftrightarrow~ x\succeq y ~\text{ but not }~ y\succeq x.$$</p>

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

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

<p>$(\Leftarrow)$ 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 $y\succeq x$.</p>

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

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

CC BY-SA 3.0
2017-09-10T19:52:42.427Z
2017-09-10T21:28:37.323Z
2017-09-10T21:28:37.323Z
Ziwei Wang
<p>Let $\succeq$ be a preference relation on $X$. Is it true that $x \succeq y$ if and only if $\lnot (y \succ x)$? </p>

<p>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 $x \succeq y$. </p>

<p>Is my proof correct? Also, in general, I just want to ask, is it true that whenever we have a <strong>preference</strong> 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 $x&gt;y$ or $y&gt;x$.</p>

CC BY-SA 3.0
2017-09-04T12:16:46.137Z
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](https://economics.stackexchange.com/questions/9385/properties-of-orders-and-preference-relations?rq=1)?
luchonacho
CC BY-SA 3.0
2017-09-04T13:40:30.177Z
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.
Kitsune Cavalry
CC BY-SA 3.0
2017-09-08T04:07:34.293Z
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.
Herr K.
CC BY-SA 3.0
2017-09-08T04:11:23.257Z
@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 $\lnot(y\succsim x)\text{ or }x\succsim y$.
Herr K.
<mathematical-economics><preferences>
Properties of preference relation
user40333
# Properties of preference relation

(Asked by: username_0 on 2017-09-04)

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 $y>x$.

**Tags:** mathematical-economics, preferences
> **Comments:**
> **username_1**: 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](https://economics.stackexchange.com/questions/9385/properties-of-orders-and-preference-relations?rq=1)?  
> **username_2**: 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.  
> **username_3**: 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.  
> **username_3**: @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 $\lnot(y\succsim x)\text{ or }x\succsim y$.  

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## 1 answers

### Answer by username_4 on 2017-09-10. Score: 2 | Accepted answer

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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