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I must proof:

$\forall x : (x \in \operatorname{field}(R) \leftrightarrow [x]_R\neq \emptyset)$

(with $\operatorname{field}(R)=\operatorname{dom}(R) \cup \operatorname{cod}(R)$ and $R$ is symmetric relation (namely $R$ is symmetric relation over $\operatorname{field}(R)$) and $[x]_R:=\{z|(x,z)\in R\}$

Proof: \begin{align*} x \in \operatorname{field}(R) &\leftrightarrow x \in \operatorname{dom}(R) \cup \operatorname{cod}(R) \\ &\leftrightarrow x \in \operatorname{dom}(R) \vee x \in \operatorname{cod}(R) \\&\leftrightarrow \exists t:((x,t) \in R) \vee \exists s:((s,x) \in R) \\ &\leftrightarrow \exists t:((x,t) \in R) \vee \exists s:((x,s) \in R) \text{ (beacause is } R \text{ symmetric relation over } \operatorname{field}(R)) \\ &\leftrightarrow \exists t:(t \in [x]_R) \vee \exists s:(s \in [x]_R) \\ &\leftrightarrow [x]_R \neq \emptyset \vee [x]_R \neq \emptyset \\ &\leftrightarrow [x]_R \neq \emptyset \end{align*}

Is it correct?

  • 3
    Your proof is correct, but it is redundant. Indeed when you prove direction "$ \to $", you are actually proving both directions. – 2017-01-26
  • 0
    ops, you are right!! – 2017-01-26

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