intersection of antisymetric relations is antisymetric Suppose $A$ is some set, and $R$ and $S$ are relations on $A$ s.t. $R$ and $S$ are anti-symmetric. I want to prove that $R\cap S$ is anti-symmetric.
Let $a,b \in A \ $ s.t. $a\ne b$ and $(a,b)\in R\cap S$ then, $(a,b)\in R$ and $(a,b)\in S$ so $(b,a)\notin R$ and $(b,a) \notin S$ so $(b,a)\notin R\cap S$
Is this correct? I feel its weird to prove it from the direct definition of anti-symmetry ($a\sim b \wedge b\sim a \implies a=b$).

If $R$ is anti-symmetric then any subset of $R$ is anti-symmetric, also $R\cap S$.
I think the normal way of doing such things is:
If $(a,b),(b,a)\in R\cap S\implies (a,b),(b,a)\in R\implies a=b$.
