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What is wrong in the following reasoning:

Every symmetric and transitive relation is a relation of equivalence

Proof:

$x \sim y \Rightarrow y\sim x$ - becuase of symmetry

$x \sim y \wedge y\sim x \Rightarrow x\sim x$ - because of transitivity

Therefore the relation is reflexive - so it is a relation of equivalence

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  • $\begingroup$ Consider the empty relation on $\{a\}$. Or if that’s a bit too outré, consider the relation $\{\langle 0,0\rangle\}$ on $\{0,1\}$. Your argument gets you $x\sim x$ only if there actually is some $y$ such that $x\sim y$. $\endgroup$ – Brian M. Scott Oct 17 '16 at 13:32
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This argument assumes that $x$ is related to some $y$, which may not be the case.

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I would elaborate the last answer. The set of things that particular x can be relative to(and due to symmetry those are relative to x) can be Empty set. If you write your argument fully mathematically - symmetry: For any(means if there is any) x ~ y => y ~ x. As a result, you may not have anything to input in this chain of logic to establish the output, which is otherwise correct.

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