Schröder Rule/Proof 1

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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

$(1)$ iff $(2)$

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

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


We shall use the method of truth tables.

The two statements will be equivalent iff the columns under the main connectives, which is $\implies$ in each case, are identical.

Statement 1:

$\begin{array}{ccccc} ((y, z) \in A & \land & (x, y) \in B) & \implies & (x, z) \in C \\ \hline T & T & T & T & T \\ T & T & T & F & F \\ T & F & F & T & T \\ T & F & F & T & F \\ F & F & T & T & T \\ F & F & T & T & F \\ F & F & F & T & T \\ F & F & F & T & F \\ \end{array}$

Statement 2:

$\begin{array}{ccccc} ((y, z) \in A & \land & (x, z) \notin C) & \implies & (x, y) \notin B \\ \hline T & F & F & T & F \\ T & T & T & F & F \\ T & F & F & T & T \\ T & T & T & T & T \\ F & F & F & T & F \\ F & F & T & T & F \\ F & F & F & T & T \\ F & F & T & T & T \\ \end{array}$

$\Box$

$(2)$ iff $(3)$



Source of Name

This entry was named for Ernst Schröder.


Sources