Schröder Rule/Proof 2

Theorem

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

The following statements are equivalent:

$(1): \quad A \circ B \subseteq C$

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

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

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:

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

This article, or a section of it, needs explaining. In particular: Actually, that only gets us to $\forall x, z \in S: \paren {\paren {\exists y \in S: \tuple {y, z} \in A \land \tuple {x, y} \in B} \implies \tuple {x, z} \in C}$. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code.

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:

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

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

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

This needs considerable tedious hard slog to complete it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Finish}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.