# Trick Proof: relation of equivalence

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

• 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 $x\sim y$. – Brian M. Scott Oct 17 '16 at 13:32

This argument assumes that $x$ is related to some $y$, which may not be the case.