Prove that a simple ordering implies a partial ordering

Question

I am having difficulty proving theorem 55 section 3.2 page 73 of Patrick Suppes Axiomatic Set Theory. This is the theorem to be proven. (R is a simple ordering) ⇒ (R is a partial ordering). He states that this theorem is obvious but this is only the case for the antisymmetry and transitivity of R. Definition of a simple ordering R is [(R is antisymmetric in FA) ● (R is transitive in FA) ● (R is strongly connected in FA)]. Definition of a partial ordering R is [(R is reflexive in FA) ● (R is transitive in FA) ● (R is antisymmetric in FA)]. So clearly all that needs to be proven is that (R is reflexive in FA). How do you do this? I tried a conditional proof. Please answer in detail.

Answer 1

Suppose that $R$ is a simple ordering (antisymmetric, transitive and strongly connected: $x,y \in A \rightarrow xRy \lor yRx$). To see that $R$ is a partial ordering, we already have transitive and antisymmetric. So we need only reflexiveness. Well, strongly connected with $y = x$ gives immediately that $xRx$ for all $x$. Done.

Answer 2

Suppose $x\in A$. Because $R$ is strongly connected, either $xRx$ or $xRx$. That is, $R$ is reflexive.