Does the topology of Hasse diagrams describe posets uniquely (i.e. what about equality)?

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Consider the set $S=\{a, b, c\}$. Now consider the following two partial orderings:

\begin{matrix} \require{color} \def\hlcheck{\colorbox{yellow}{$\checkmark$}} \leq_1 &a &b &c \\ a &\checkmark &\hlcheck &\checkmark \\ b &\hlcheck &\checkmark &\checkmark \\ c &\times &\times &\checkmark \\ \end{matrix}

 

\begin{matrix} \require{color} \def\hltimes{\colorbox{yellow}{$\times$}} \leq_2 &a &b &c \\ a &\checkmark &\hltimes &\checkmark \\ b &\hltimes &\checkmark &\checkmark \\ c &\times &\times &\checkmark \\ \end{matrix}

Note that the difference between $\leq_1$ and $\leq_2$ is that in the former, $a=b$, while in the latter $a$ and $b$ are not ordered with respect to each other.

It seems to me that the following Hasse diagram represents both the partially ordered sets $\{S, \leq_1\}$ and $\{S, \leq_2\}$. Is this true? If so, is there any standard way to differentiate between both partial orderings?

c
|\
a b
1

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No, it isn't true.
The difference is that in $\leq_1$, you have $a=b$ because of anti-symmetry (so $a$ and $b$ are only one element).