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?
