<p>Probably it can be done easier if you do both steps separately (<span class="math-container">$\implies$</span> and <span class="math-container">$\impliedby$</span>), but here is a proof that does both at the same time:
<span class="math-container">\begin{align*}
&amp;x\succ y \vee x\sim y\\
\iff\;&amp; x \succ y \vee (x\succeq y \wedge y \succeq x)  &amp; \text{Definition of $\sim$}   \\
\iff\;&amp; (x \succ y \vee x\succeq y) \wedge (x \succ y \vee y \succeq x)&amp; \text{Distributivity}\\
\iff\;&amp; [x \succ y \vee \neg( y \succ x)] \wedge [x \succ y \vee \neg(x\succ y)]&amp; \text{Definition  of $\succeq$}\\
\iff\;&amp; x \succ y \vee \neg( y \succ x)&amp; \text{LEM}\\
\iff\;&amp; [x \succ y \vee \neg( y \succ x)] \wedge[x \succ y \to \neg( y \succ x)]&amp; \text{Asymmetry}\\
\iff\;&amp; [x \succ y \vee \neg( y \succ x)] \wedge[\neg(x \succ y) \vee \neg( y \succ x)]&amp; \text{Definition of $\to$}\\
\iff\;&amp; [x \succ y \wedge \neg(x \succ y)] \vee \neg( y \succ x)&amp; \text{Distributivity}\\
\iff\;&amp; \neg( y \succ x)&amp; \text{Contradiction}\\
\iff\;&amp; x \succeq y&amp; \text{Definition  of $\succeq$}\\
\end{align*}</span></p>
<p>Negative transitivity seems to not be a necessary condition for this proposition.</p>

CC BY-SA 4.0
2021-12-19T12:18:14.417Z
2021-12-19T15:27:26.147Z
2021-12-19T15:27:26.147Z
LudwigNagasena
<p>Let  <span class="math-container">$ \succ $</span> be a binary relationship on the set <span class="math-container">$X$</span> such that, given any <span class="math-container">$ x, y, z\in X $</span>, if <span class="math-container">$x\succ y$</span>:</p>
<ol>
<li>(Asymmetry):  <span class="math-container">$\neg(y\succ x)$</span>,</li>
<li>(Negative transitivity): <span class="math-container">$(x\succ z) \vee (z\succ y)$</span>.</li>
</ol>
<p>Let us define the abbreviations:</p>
<ol start="3">
<li><p><span class="math-container">$x\succeq y \;:=\; \neg(y\succ x)  $</span>,</p>
</li>
<li><p><span class="math-container">$x \sim y \;:=\; x\succeq y\; \wedge \;y \succeq x$</span>.</p>
</li>
</ol>
<p>As usual, the relations <span class="math-container">$\succ, \succeq, \sim$</span> denote strong preference, weak preference, and  indifference.</p>
<p>Intuition suggests that I can conclude:
<span class="math-container">$$x\succeq y \; \leftrightarrow \;(x\succ y\; \vee \;x\sim y) $$</span></p>
<p>If so, how can I derive it formally? Any useful references?</p>

CC BY-SA 4.0
2021-12-19T03:39:07.090Z
Is the relation $\succeq$ complete?
Giskard
CC BY-SA 4.0
2021-12-19T08:09:14.587Z
@Giskard:     The definitions above are taken from Föllmer, Schied, Stochastic Finance, where it is stated that $\succeq$ completeness is implied by the asymmetry and the negative transitivity of $\succ$.
antonio
CC BY-SA 4.0
2021-12-19T08:44:56.713Z
Thanks for the source. Remark 2.3. claims that these properties are equivalent to $\succeq$ being transitive and complete. Can you then not take the usual $\succeq$ route to prove your intuitited statement?
Giskard
CC BY-SA 4.0
2021-12-19T09:10:28.220Z
@Giskard: Well, usually  $x\succ y\; \vee \;x\sim y$ is taken as the definition of $ x\succeq y$.
antonio
<preferences><choice-theory><proof><proof-reference>
Weak preferences and negative transitivity
antonio
# Weak preferences and negative transitivity

(Asked by: username_0 on 2021-12-19)

Let $ \succ $ be a binary relationship on the set $X$ such that, given any $ x, y, z\in X $, if $x\succ y$:

1. (Asymmetry): $\neg(y\succ x)$,
2. (Negative transitivity): $(x\succ z) \vee (z\succ y)$.

Let us define the abbreviations:

3. $x\succeq y \;:=\; \neg(y\succ x) $,
4. $x \sim y \;:=\; x\succeq y\; \wedge \;y \succeq x$.

As usual, the relations $\succ, \succeq, \sim$ denote strong preference, weak preference, and indifference.

Intuition suggests that I can conclude:
$$x\succeq y \; \leftrightarrow \;(x\succ y\; \vee \;x\sim y) $$

If so, how can I derive it formally? Any useful references?

**Tags:** preferences, choice-theory, proof, proof-reference

> **Comments:**
> **username_1**: Is the relation $\succeq$ complete?  
> **username_0**: @username_1:     The definitions above are taken from Föllmer, Schied, Stochastic Finance, where it is stated that $\succeq$ completeness is implied by the asymmetry and the negative transitivity of $\succ$.  
> **username_1**: Thanks for the source. Remark 2.3. claims that these properties are equivalent to $\succeq$ being transitive and complete. Can you then not take the usual $\succeq$ route to prove your intuitited statement?  
> **username_0**: @username_1: Well, usually  $x\succ y\; \vee \;x\sim y$ is taken as the definition of $ x\succeq y$.  

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## 1 answers

### Answer by username_2 on 2021-12-19. Score: 5 | Accepted answer

Probably it can be done easier if you do both steps separately ($\implies$ and $\impliedby$), but here is a proof that does both at the same time:
\begin{align\*}
&x\succ y \vee x\sim y\\
\iff\;& x \succ y \vee (x\succeq y \wedge y \succeq x) & \text{Definition of $\sim$} \\
\iff\;& (x \succ y \vee x\succeq y) \wedge (x \succ y \vee y \succeq x)& \text{Distributivity}\\
\iff\;& [x \succ y \vee \neg( y \succ x)] \wedge [x \succ y \vee \neg(x\succ y)]& \text{Definition of $\succeq$}\\
\iff\;& x \succ y \vee \neg( y \succ x)& \text{LEM}\\
\iff\;& [x \succ y \vee \neg( y \succ x)] \wedge[x \succ y \to \neg( y \succ x)]& \text{Asymmetry}\\
\iff\;& [x \succ y \vee \neg( y \succ x)] \wedge[\neg(x \succ y) \vee \neg( y \succ x)]& \text{Definition of $\to$}\\
\iff\;& [x \succ y \wedge \neg(x \succ y)] \vee \neg( y \succ x)& \text{Distributivity}\\
\iff\;& \neg( y \succ x)& \text{Contradiction}\\
\iff\;& x \succeq y& \text{Definition of $\succeq$}\\
\end{align\*}

Negative transitivity seems to not be a necessary condition for this proposition.


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