Denial of Existence/Examples/x less than or equal to 3/Examples/0 to 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
- $\lnot P$ is false
The truth of $P$ can be demonstrated by citing $x \in S: x = 2$.
Thus $2$ is an instance of an $x \in S$ such that $x > 3$ that has been shown to exist.