Relative Complement of Cartesian Product/Proof 2

Theorem

Let $A$ and $B$ be sets.

Let $X \subseteq A$ and $Y \subseteq B$.

Then:

$\relcomp {A \mathop \times B} {X \times Y} = \paren {A \times \relcomp B Y} \cup \paren {\relcomp A X \times B}$

Proof

\(\ds \)

\(\)

\(\ds \relcomp {A \mathop \times B} {X \times Y}\)

\(\ds \leadstoandfrom \ \ \)

\(\ds \)

\(\)

\(\ds A \times B \setminus X \times Y\)

Definition of Relative Complement

\(\ds \leadstoandfrom \ \ \)

\(\ds \)

\(\)

\(\ds \set {\tuple {x, y}: x \in A \land y \in B \land \neg \paren {x \in X \land y \in Y} }\)

Definition of Cartesian Product

\(\ds \leadstoandfrom \ \ \)

\(\ds \)

\(\)

\(\ds \set {\tuple {x, y}: x \in A \land y \in B \land \paren {x \notin X \lor y \notin Y} }\)

De Morgan's Laws: Disjunction of Negations

\(\ds \leadstoandfrom \ \ \)

\(\ds \)

\(\)

\(\ds \paren {A \times \relcomp B Y} \cup \paren {\relcomp A X \times B}\)

Definition of Cartesian Product

and so by definition of set equality:

$\relcomp {A \mathop \times B} {X \times Y} = \paren {A \times \relcomp B Y} \cup \paren {\relcomp A X \times B}$

$\blacksquare$