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:

  1. $x\succeq y \;:=\; \neg(y\succ x) $,

  2. $x \sim y \;:=\; x\succeq y\; \wedge \;y \succeq x$.

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?

  • $\begingroup$ Is the relation $\succeq$ complete? $\endgroup$
    – Giskard
    Commented Dec 19, 2021 at 3:39
  • $\begingroup$ @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 $\succ$. $\endgroup$
    – antonio
    Commented Dec 19, 2021 at 8:09
  • $\begingroup$ 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? $\endgroup$
    – Giskard
    Commented Dec 19, 2021 at 8:44
  • $\begingroup$ @Giskard: Well, usually $x\succ y\; \vee \;x\sim y$ is taken as the definition of $ x\succeq y$. $\endgroup$
    – antonio
    Commented Dec 19, 2021 at 9:10

1 Answer 1


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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