how to express does not exist by quantifiers

Question

Hi guys I have a problem with logic, look at this phrase:

"aucun entier n'est supérieur a tous les entiers" (No integer is superior to all integers)

I need to write it on mathematical form.

"there does not exist any ... such that..." is the same as "for all ... holds the negation of ..." — Max, Oct 21 '15 at 09:54
gyes can same one help my http://math.stackexchange.com/questions/1490543/how-to-express-this-phrases-by-quantifiers — , Oct 21 '15 at 12:14

score 2 · Accepted Answer · answered Oct 21 '15 at 09:55

2

$$\nexists n\in\mathbb{Z}: \forall m\in\mathbb{Z}, m<n$$ or $$\forall n\in\mathbb{Z}: \exists m\in\mathbb{Z}, n<m$$

answered Oct 21 '15 at 09:55

Zelos Malum

( thier is no ∄ in math symbole that the problime ) – Khalid Es-safi Oct 21 '15 at 09:58
that is why I gave the alternative, they both mean the same :) – Zelos Malum Oct 21 '15 at 10:00
no is note mean the same this(∀n∈Z:∃m∈Z,n<m) is négation of (∄n∈Z:∀m∈Z,m<n) that mean if (∄n∈Z:∀m∈Z,m<n ) True the négation is incorrect – Khalid Es-safi Oct 21 '15 at 10:04
The first one says "there is no greatröest integer" the latter says "for all integers, there is a greater integer". Those statenents are equivalent – Zelos Malum Oct 21 '15 at 10:06
@ZelosMalum Yes, that's because they both happen to be true. But formally if you replace $\mathbb Z$ with a finite set of integer then it's no longer true, because the second will be false. – skyking Oct 21 '15 at 10:09
That is correct but we are dealing with integers so it is the entire set – Zelos Malum Oct 21 '15 at 10:10
@ZelosMalum The point is that they are two correct, but different facts about the integers. You could as well have claimed that all integers are either odd or even and even that would have been equivalent (since it too is true). – skyking Oct 21 '15 at 10:14

score 1 · Answer 2 · answered Oct 21 '15 at 10:03

1

If $\nexists$ wont do you can also use $\neg\exists$ instead:

$$\nexists n\in\mathbb N:\forall m\in\mathbb N: m<n$$ or

$$\neg\exists n\in\mathbb N:\forall m\in\mathbb N: m<n$$

Or one could equivalently use De Morgan rule:

$$\forall n\in\mathbb N:\exists m\in\mathbb N: m\ge n$$

even though it's not formulated differently.

answered Oct 21 '15 at 10:03

skyking

but (¬∃) is the négation of ∃ and it like ∀ look (∀=¬∃) – Khalid Es-safi Oct 21 '15 at 10:06
@KhalidEs-safi No, $\exists$ is not a complete statement, the negation refers to the complete statement. $\forall$ is defenitely no the same as $\neg\exists$, the De Morgan law says that $\neg\exists$ is the same as $\forall\neg$ and $\neg\forall$ is the same as $\exists\neg$ (but it has to continue into a complete statement). – skyking Oct 21 '15 at 10:11

2 Answers2