# Denial of Existence/Examples/x less than or equal to 3

## Contents

## Examples of Denial of Existence

Let $S \subseteq \R$ be a subset of the real numbers.

Let $P$ be the statement:

- $\exists x \in S: x \le 3$

The negation of $P$ is the statement written in its simplest form as:

- $\forall x \in S: x > 3$

## Proof

\(\displaystyle \) | \(\) | \(\displaystyle \lnot \exists x \in S: x \le 3\) | |||||||||||

\(\displaystyle \) | \(\leadsto\) | \(\displaystyle \forall x \in S: \lnot x \le 3\) | Denial of Universality | ||||||||||

\(\displaystyle \) | \(\leadsto\) | \(\displaystyle \forall x \in S: x \not \le 3\) | |||||||||||

\(\displaystyle \) | \(\leadsto\) | \(\displaystyle \forall x \in S: x > 3\) |

$\blacksquare$

## Examples

### Example where $S = \set {2, 3, 4}$

Let $P$ be the statement:

- $\exists x \in S: x \le 3$

and $\lnot P$ its negation:

- $\forall x \in S: x > 3$

Let $S = \set {2, 3, 4}$.

Then we have that:

- $P$ is true

and consequently:

- $\lnot P$ is false

### Example where $S = \closedint 0 3$

Let $P$ be the statement:

- $\exists x \in S: x \le 3$

and $\lnot P$ its negation:

- $\forall x \in S: x > 3$

Let $S = \closedint 0 3$ where $\closedint \cdot \cdot$ denotes a closed real interval.

Then we have that:

- $P$ is true

and consequently:

- $\lnot P$ is false

## Sources

- 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): Chapter $1$: Sets and Logic: Exercise $1 \ \text{(ii)}$