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