$(∀x)(∃y)(x>y)$ is false. But then why is $ (∀x)(∃y)(x\geq y)$ true?

Question

Given the Universe is the set of natural numbers, then $(∀x)(∃y)(x>y)$ is false. But then why is $(∀x)(∃y)(x\geq y)$ true?

The first equation and the second equation is the same except for "=" in the second equation. so i don't get how that affects the statement to this degree. Thank you.

Answer 1

Look at the smallest number in the natural numbers $\mathbb N$ (since you mention that the universe is the set of all natural numbers): the smallest number in $\mathbb N$ would be $0$ or $1$, depending on your definition. Let's just go with $0$.

Since $0 \in \mathbb N$, the universally quantified "x" means the inequality must hold for every natural number $x$, including $x = 0.\;$ Now, does there exist any $y \in \mathbb N$ such that that $0 > y\;$?

On the other hand, if we allow equality too, then we have that it is true that there exists a $\,y \in \mathbb N\,$ such that $\;0 \geq y,\,$ namely, $\,y = 0:$ That is, it is certainly true that $\;0 \geq 0.$

The same logic applies if $x = 1$ instead of $x = 0$, if you are working with a definition of the natural numbers $\{1, 2, 3, \ldots\}$.

Answer 2

Any ordered set with a minimum element provides an example where the first is false and the second is true, because there can be nothing below the minimum element.

Answer 3

Another take on this question is to try and write the latter in terms of the former: \begin{align} & \langle \forall x :: \langle \exists y :: x \geq y \rangle \rangle \\ \equiv & \;\;\;\;\;\text{"write $\;\geq\;$ in terms of $\;>\;$ -- since know something about that already"} \\ & \langle \forall x :: \langle \exists y :: x > y \:\lor\: x = y \rangle \rangle \\ \equiv & \;\;\;\;\;\text{"logic: $\;\exists\;$ distributes over $\;\lor\;$"} \\ & \langle \forall x :: \langle \exists y :: x > y \rangle \:\lor\: \langle \exists y :: x = y \rangle \rangle \\ \equiv & \;\;\;\;\;\text{"logic: one-point rule"} \\ & \langle \forall x :: \langle \exists y :: x > y \rangle \:\lor\: \textrm{true} \rangle \\ \equiv & \;\;\;\;\;\text{"logic: simplify using $\;P \lor \text{true} \equiv \text{true}\;$"} \\ & \text{true} \\ \end{align} Now we see that we didn't even use the fact that $\;\langle \forall x :: \langle \exists y :: x > y \rangle \rangle\;$ is false on $\;\mathbb N\;$...