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:


 * $\forall x \in S: \tuple {x, x} \in S \times S$

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

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


 * $\forall x, y \in S: \tuple {x, y} \in S \times S \land \tuple {y, x} \in S \times S$

Since we have by definition of Cartesian product that:


 * $\forall x, y \in S: \tuple {y, x} \in S \times S$

this follows by True Statement is implied by Every Statement.

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


 * $\tuple {x, y} \in S \times S \land \tuple {y, z} \in S \times S \implies \tuple {x, z} \in S \times S$

By definition of Cartesian product, we have that:


 * $\forall x, z \in S: \tuple {x, z} \in S \times S$

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.