Schröder Rule/Proof 2

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


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

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


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

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


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

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


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

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


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

So in all we have:
 * $(1') \quad \forall x, y, z \in S: \paren {\tuple {y, z} \in A \land \tuple {x, y} \in B \implies \tuple {x, z} \in C}$
 * $(2'') \quad \forall x, y, z \in S: \paren {\tuple {y, z} \in A \land \tuple {x, z} \notin C \implies \tuple {x, y} \notin B}$
 * $(3'') \quad \forall x, y, z \in S: \paren {\tuple {x, z} \notin C \land \tuple {x, y} \in B \implies \tuple {y, z} \notin A}$