# Schröder Rule/Proof 2

## Theorem

Let $A$, $B$ and $C$ be relations on a set $S$.

Then the following are equivalent statements:

$(1): \quad A \circ B \subseteq C$
$(2): \quad A^{-1} \circ \overline{C} \subseteq \overline{B}$
$(3): \quad \overline{C} \circ B^{-1} \subseteq \overline{A}$

where:

$\circ$ denotes relation composition
$A^{-1}$ denotes the inverse of $A$
$\overline{A}$ denotes the complement of $A$.

## Proof

By the definition of relation composition and subset we have that statement $(1)$ may be written as:

$(1')\quad \forall x, y, z \in S: \left({ (y, z) \in A \land (x, y) \in B \implies (x, z) \in C }\right)$

Using a different arrangement of variable names, statement $(2)$ can be written:

$(2')\quad \forall x, y, z \in S: \left({ (z, y) \in A^{-1} \land (x, z) \in \overline C \implies (x, y) \in \overline B }\right)$

By the definition of the inverse and the complement of a relation we can rewrite this as:

$(2'')\quad \forall x, y, z \in S: \left({ (y, z) \in A \land (x, z) \notin C) \implies (x, y) \notin B }\right)$

Similarly, statement $(3)$ can be written:

$(3')\quad \forall x, y, z \in S: \left({ (x, z) \in \overline C \land (y, x) \in B^{-1} \implies (y, z) \in \overline A }\right)$

By the definition of the inverse and the complement of a relation we can rewrite this as:

$(3'')\quad \forall x, y, z \in S: \left({ (x, z) \notin C \land (x, y) \in B \implies (y, z) \notin A }\right)$

So in all we have:

$(1')\quad \forall x, y, z \in S: \left({ (y, z) \in A \land (x, y) \in B \implies (x, z) \in C }\right)$
$(2'')\quad \forall x, y, z \in S: \left({ (y, z) \in A \land (x, z) \notin C) \implies (x, y) \notin B }\right)$
$(3'')\quad \forall x, y, z \in S: \left({ (x, z) \notin C \land (x, y) \in B \implies (y, z) \notin A }\right)$