Suppose $R$ is a partial order on $A$ and $B \subseteq A$. Prove that $R \cap (B \times B)$ a partial order on $B$. Can somebody show me how to prove this? I would much appreciate it if one could show the givens and goals similar to how it is set out in Velleman's 'how to prove it' book, though any help would be good. Thanks.
 A: I don't know Velleman's book, but I support your seek for guidance. So let me try to provide you with a structure that hopefully mimics the one you are looking for.
Givens: We have a partial order $R$ on $A$ and some subset $B \subseteq A$. So $A$ is a (nonempty) set, $R \subseteq A \times A$ is a binary relation on $A$ which satisfies


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*Reflexivity: For all $a \in A$, we have $(a,a) \in R$ (- often authors use an infix notation for binary relations - in this case replace $(x,y) \in R$ with $x R y$ throughout my entire post).

*Antisymmetry: For all $a, b \in A$, if $(a,b) \in R$ and $(b,a) \in R$, then $a = b$.

*Transitivity: For all $a, b, c \in A$, if $(a,b) \in R$ and $(b,c) \in R$, then $(a,c) \in R$.


Goals: We want to show that $S := R \cap (B \times B)$ is a partial order on $B$. Since $S$ clearly is a binary relation on $B$ (i.e. $S \subseteq B \times B$), it suffices to verify


*

*Reflexivity: For all $a \in B$, we have $(a,a) \in S$.

*Antisymmetry: For all $a, b \in B$, if $(a,b) \in S$ and $(b,a) \in S$, then $a = b$.

*Transitivity: For all $a, b, c \in B$, if $(a,b) \in S$ and $(b,c) \in S$, then $(a,c) \in S$.
