To prove that a relation $R$ on a set $A$ is symmetric if and only if $R^{-1} = R$, let us proceed with the proof in two directions.

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Part 1: If $R$ is symmetric, then $R^{-1} = R$

Assumption:

The relation $R$ is symmetric. By definition of symmetry, this means: $(x, y) \in R \implies (y, x) \in R \quad \text{for all } x, y \in A.$

Goal:

Show that $R^{-1} = R$.

Proof:

1. By definition, the inverse relation $R^{-1}$ is: $R^{-1} = \{(y, x) \mid (x, y) \in R\}.$
2. Since $R$ is symmetric, for every $(x, y) \in R$, we also have $(y, x) \in R$.
3. Thus, every element in $R^{-1}$ (which is $(y, x)$ for some $(x, y) \in R$) is also in $R$. Therefore, $R^{-1} \subseteq R$.
4. Conversely, every element in $R$ (which is $(x, y)$) has its reverse $(y, x)$ also in $R$ due to symmetry. Hence, every element in $R$ is in $R^{-1}$, so $R \subseteq R^{-1}$.
5. Since $R^{-1} \subseteq R$ and $R \subseteq R^{-1}$, we conclude that: $R^{-1} = R.$

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Part 2: If $R^{-1} = R$, then $R$ is symmetric

Assumption:

The relation $R^{-1} = R$.

Goal:

Show that $R$ is symmetric.

Proof:

1. By definition, $R^{-1} = \{(y, x) \mid (x, y) \in R\}$.
2. Given $R^{-1} = R$, this means: $(y, x) \in R \iff (x, y) \in R \quad \text{for all } x, y \in A.$
3. This is precisely the definition of symmetry for $R$.
4. Therefore, $R$ is symmetric.

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Conclusion

We have proven both directions:

1. If $R$ is symmetric, then $R^{-1} = R$.
2. If $R^{-1} = R$, then $R$ is symmetric.

Thus: $R \text{ is symmetric if and only if } R^{-1} = R.$

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Do you want further clarification, examples, or additional proofs?

Here are five follow-up questions you may consider:

1. Can you provide an example of a symmetric relation on a set and its inverse?
2. What changes if the relation is reflexive but not symmetric?
3. How does the proof change for relations on infinite sets?
4. What are examples of relations where $R^{-1} \neq R$, and why?
5. How is this property used in graph theory or matrix representations of relations?

Tip: Always verify symmetry by explicitly checking the pair $(y, x)$ for every $(x, y)$ in the relation.