Let
and
be equivalent, and
.
Then
,
and, by transitivity, also
,
thus
.
Therefore, the equivalence classes coincide. The implication from the middle to the right is clear, because, due to
,
the equivalence classes are not empty. Suppose now that
,
and let denote an element in the intersection. Then
and ,
and, by transitivity,
.
Because of the reflexivity, we have
;
therefore,
.
This union is disjoint by part (1).
The surjectivity is clear because of the definition of the quotient set, and since is sent to the class .
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![{\displaystyle {}x\in [x]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6e12337a925c460804120489ba31028dd0fadeff)
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