Does the definition of a partial order require the definition of an equivalence relation?

Question

Consider the set $S$ and a binary relation $R \subseteq S \times S$. $R$ is a partial order on $S$ if

1. $\forall s \in S, (s,s) \in R$ (reflexivity)
2. $\forall s_1,s_2 \in S, (s_1,s_2) \in R \wedge (s_2,s_1) \in R \implies s_1 = s_2$ (antisymmetry)
3. $\forall s_1,s_2,s_3 \in S, (s_1,s_2) \in R \wedge (s_2,s_3) \in R \implies (s_1,s_3) \in R$ (transitivity)

My question is about the antisymmetry property in (2). It seems to me that the statement "$s_1 = s_2$" can only be evaluated to be true or false if there already exists an equivalence relation $\sim \, \subset S \times S$. That is, $s_1 = s_2 \iff (s_1,s_2) \in \, \sim$. So, we cannot define a partial order on $S$ without first defining an equivalence relation on $S$ to check the antisymmetry property. Is my understanding correct?