Trivial Relation is Equivalence

Theorem
The trivial relation on $S$:


 * $\mathcal R = S \times S$

is always an equivalence in $S$.

Proof
Let us verify the conditions for an equivalence in turn.

Reflexivity
For $\mathcal R$ to be reflexive means:


 * $\forall x \in S: \left({x, x}\right) \in S \times S$

which is trivial by definition of the Cartesian product $S \times S$.

Symmetry
For $\mathcal R$ to be symmetric means:


 * $\forall x, y \in S: \left({x, y}\right) \in S \times S \land \left({y, x}\right) \in S \times S$

Since we have by definition of Cartesian product that:


 * $\forall x, y \in S: \left({y, x}\right) \in S \times S$

this follows by True Statement is Implied by Every Statement.

Transitivity
For $\mathcal R$ to be transitive means:


 * $\left({x, y}\right) \in S \times S \land \left({y, z}\right) \in S \times S \implies \left({x, z}\right) \in S \times S$

By definition of Cartesian product, we have that:


 * $\forall x, z \in S: \left({x, z}\right) \in S \times S$

hence by True Statement is Implied by Every Statement, it follows that $\mathcal R$ is transitive.

Having verified all three conditions, we conclude $\mathcal R$ is an equivalence.