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Prove if $\succsim$ is rational then: if $x \succ y \succsim z$, then $x \succ z$

By definition of $\succ$,

\begin{equation} \tag{1} x \succ y \iff x \succsim y, \; \neg \; y \succsim x \end{equation}

Where $\neg$ is the negation symbol.

So we want to show both

$$ x \succsim z \; and \; \neg z \succsim x $$

We are given

\begin{equation} \tag{2} y \succsim z \end{equation}

By $(1)$ and $(2)$ and transitivity of $\succsim$ (since $\succsim$ is rational),

\begin{equation} \tag{3} x \succsim y \succsim z \Rightarrow x \succsim z \end{equation}

We also need $\neg \; z \succsim x$

Suppose the contrary that $z \succsim x$, but then

$$ z \succsim x \succ y \succsim z $$

Such that

\begin{equation} \tag{4} z \succ z \end{equation}

is a contradiction (as $\succ$ is irreflexive) so it must be that

\begin{equation} \tag{5} \neg \; z \succsim x \end{equation}

Applying $(3)$ and $(5)$ we have the desired property

$$ x \succsim z, \; \neg \; z \succsim x \Rightarrow x \succ z $$

My question is if I am allowed to arrive at the statement $(4)$ from the line above it?

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  • $\begingroup$ No. Getting to $(4)$ the way you do requires the sort of "transitivity" on $\succ$ that you are trying to show in the first place. $\endgroup$ – Kitsune Cavalry Sep 28 '16 at 22:12
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By contradiction, assume $z \succsim x$. However, we already know that $y \succsim z$, which is a contradiction (after a couple more steps). Can you now see this?

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  • $\begingroup$ Hi @Jay, thank you, yes I see how it is proved this way; I was wondering if we could prove it quicker another way which Kitsune Cavalry answered. $\endgroup$ – Sunhwa Sep 29 '16 at 12:02

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