Reflexive Reduction is Antireflexive

Theorem
Let $\mathcal R$ be a relation on a set $S$.

Then $\mathcal R^\ne$, the reflexive reduction of $\mathcal R$, is antireflexive.

Proof
By the definition of reflexive reduction:


 * $\mathcal R^\neq = \mathcal R \setminus \Delta_S$

where $\Delta_S$ denotes the diagonal relation on $S$.

By Set Difference Intersection Second Set is Empty Set:


 * $\left({\mathcal R \setminus \Delta_S}\right) \cap \Delta_S = \varnothing$

Thus $\mathcal R \setminus \Delta_S$ and $\Delta_S$ are disjoint.

Hence by Antireflexive Disjoint from Diagonal Relation, $\mathcal R^\neq$ is antireflexive.