Complement of Vertical Section of Set is Vertical Section of Complement

Theorem
Let $X$ and $Y$ be sets.

Let $E \subseteq X \times Y$.

Let $x \in X$.

Then:


 * $\paren {\paren {X \times Y} \setminus E}_x = Y \setminus E_x$

where:


 * $\paren {\paren {X \times Y} \setminus E}_x$ is the $x$-vertical section of the set difference $\paren {X \times Y} \setminus E$
 * $E_x$ is the $x$-vertical section of $E$.

Proof
Note that from the definition of set difference, we have that:


 * $y \in Y \setminus E_x$




 * $y \in Y$ and $y \not \in E_x$.

That is, from the definition of the $x$-vertical section:


 * $y \in Y$ and $\tuple {x, y} \not \in E$.

This is equivalent to:


 * $\tuple {x, y} \in \paren {X \times Y} \setminus E$

From the definition of the $x$-vertical section, this is then equivalent to:


 * $y \in \paren {\paren {X \times Y} \setminus E}_x$

So we have:


 * $y \in Y \setminus E_x$ $y \in \paren {\paren {X \times Y} \setminus E}_x$.

So:


 * $\paren {\paren {X \times Y} \setminus E}_x = Y \setminus E_x$