A binary relation $\mathcal R$ on $E$ with graph $G$ is symmetric if and only if $\forall(x,y)\in E^2,\,x\mathcal R y\implies y\mathcal R x$.
Suppose that $\mathcal R$ is symmetric. By definition, $G\subset E^2$. So, if I take a random element $(x,y)$ of $E^2$, and find that it is in $G$, then I would know that $(y,x)$ is also in $G$. But if I happen to fall on $(y,x)$ first, wouldn't I conclude that $(x,y)\in G$, too? Which means that the definition should have an equivalence symbol, no?
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$\begingroup$No. The definition is what you wrote, and then you deduced from it that$$\left(\forall (x,y)\in E^2\right):x\mathrel{\mathcal R}y\iff y\mathrel{\mathcal R}x.\tag1$$But that doesn't make the definition wrong somehow. It only means that if we had decided to adopt $(1)$ has the definition, nothing would change.
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