If $R_1$ and $R_2$ are symmetric then is $R_1 - R_2$ are symmetric

Question

If $R_1$ and $R_2$ are symmetric relations on set A then is $R_1 - R_2$ are symmetric

So we have $R_1 = (R_{1})^{-1}$ and $R_2 = (R_{2})^{-1}$

Let $(a,b) \in R_1 - R_2$ . So $(a,b) \in R_1$ and $(a,b) \notin R_2$

So $(a,b) \in (R_{1})^{-1}$ and $(a,b) \notin (R_{2})^{-1}$.

So $(b,a) \in R_1$ and $(b,a) \notin R_2$. So $(b,a) \in R_1 - R_2$. Hence $(a,b) \in (R_1 - R_2)^{-1}$

Is this correct ?

Thanks.

Answer 1

Indeed.

We are given the premise that $R_1, R_2$ are symmetric relations.   Now, any element in $R_1\smallsetminus R_2$, say $(a,b)$ , is an element in $R_1$ but not in $R_2$, and vice versa; by definition of set difference.   Since $R_1$ is symmetric, therefore $(b,a)$ is in $R_1$ when $(a,b)$ is.   Since $R_2$ is symmetric, therefore $(b,a)$ is not in $R_2$ when $(a,b)$ is not.   So, we have that: if any $(a,b)$ is in $R_1\smallsetminus R_2$ then $(b,a)$ is in $R_1\smallsetminus R_2$.   This is the very definition of a symmetric relation.   Therefore we conclude that $R_1\smallsetminus R_2$ is symmetric if $R_1$ and $R_2$ are symmetric relations.

$$\small\bbox[pink]{\begin{split}\forall a\forall b &: (a,b)\in R_1\smallsetminus R_2\leftrightarrow (a,b)\in R_1\wedge (a,b)\notin R_2 &\qquad&\text{Definition of Set Difference}\\ \forall a\forall b &: (a,b)\in R_1\to (b,a)\in R_1&&\text{Definition of Symmetry}\\\forall a\forall b&: (a,b)\notin R_2\to(b,a)\notin R_2&&\text{Contrapositive Definition of Symmetry}\\\forall a\forall b &: (b,a)\in R_1\wedge (b,a)\notin R_2\leftrightarrow (b,a)\in R_1\smallsetminus R_2 &\qquad&\text{Definition of Set Difference}\\\hline \forall a\forall b &: (a,b)\in R_1\smallsetminus R_2\to (b,a)\in R_1\smallsetminus R_2& \therefore& R_1\smallsetminus R_2\text{ is symmetric if }R_1,R_2\text{ are.}\end{split}}$$

Answer 2

Yes. A similar way of saying it is that intersection of symmetric relations is symmetric and that $R$ is symmetric (if and) only if the complementary $R^c$ of $R$ is symmetric. Then $R_1\setminus R_2=R_2^c\cap R_1$.