Negating quantifiers or statements

Question

How do I symbolize the statement "there does not exist a $x$ for all $y$, $B(x,y)$".

Using $\neg(\exists x) (\forall y) B(x,y)$ would mean it is not the case that there exist a $x$ for all $y$, $B(x,y)$ or that the negation would be for both the quantifiers.
If we write $(\neg (\exists x))(\forall y) B(x,y)$ then what would negation of quantifier mean ? As far as I know negation is for statements and not quantifiers. Interchanging of quantifiers is also I think not allowed.

The basic question was if it is possible to negate one quantifier and not the other. Like for example how to symbolize not for all $x$ $B(x,y)$ is true for all $y$ ?

Answer 1

How do I symbolize the statement "there does not exist a $x$ for all $y$, $B(x,y)$"?

Using $\neg(\exists x) (\forall y) B(x,y)$ would mean it is not the case that there exist a $x$ for all $y$, $B(x,y),$ or that the negation would be for both the quantifiers.

If we write $(\neg (\exists x))(\forall y) B(x,y)$ then what would negation of quantifier mean ?

You need to first try to make sense of the original statement. In this case, it isn't actually meaningful unless we change it to something like

• there does not exist an $x$ for which this holds: for each $y$, $B(x,y)$

or

• there is no $x$ such that for each $y$, $B(x,y).$

Symbolically:

• $\lnot\bigg(\exists x\Big(\forall y \big(B(x,y)\big)\Big)\bigg).$

Notice that we are actually negating the entire sentence; quantifiers and variables don't attain any truth value, so, indeed, cannot be negated.

More human-friendly:

• $\lnot\,\exists x\,\forall y\, B(x,y)$
• $\lnot\,\exists x\,\forall y\, Bxy.$

P.S. Equivalently:

• $\forall x\, (\lnot\,\forall y\; Bxy)$
• $\forall x\;\exists y\,( \lnot\, Bxy).$

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Addendum to include comment under another answer

The sentence There is a lid for every pot is not translated as ∃l ∀p or as ∀p ∃l, which are not sentences, let alone meaningful. Because of its hanging quantifier, this sentence is technically ambiguous, and can actually be read as either of the following:

• ∀p ∃l (p∈P → (l∈L ∧ F(l,p))

  or, equivalently, ∀p∈P ∃l∈L F(l,p)

  Each pot has a lid

  For each pot, there is some lid such that the lid fits the pot

• ∃l ∀p (l∈L ∧ (p∈P → F(l,p))

  or, equivalently, ∃l∈L ∀p∈P F(l,p)

  Some lid fits all pots

  There is some lid such that for every pot, the lid fits the pot.

Answer 2

Let $B(x,y) : xy=0$
Let our Domain be Integers.

We can make the Claim : $$\{ \exists x : [\forall y : B(x,y)] \}\tag{1}$$ In other words ,
$$\{ \exists x : [\forall y : xy=0] \}$$

It is indeed true , we can take $x=0$ , hence there does exists some $x$ such that $xy=0$ for all $y$ which are in our Domain Integers.

Hence the Negation must be false.

Now when we change the Domain to be Irrational numbers , there is no such number $x$ (Irrational number) which makes $xy=0$ for all $y$ which are in our Domain Irrational numbers.

Hence the Negation must be true now.

In general , we do not know whether the Claim in true or not.
We just take the written Claim and try to make the Negation.

$$\lnot \{ \exists x : [\forall y : B(x,y)] \}$$ We can use the "Standard" laws : $$\{ \forall x : \lnot [\forall y : B(x,y)] \}\tag{2A}$$ $$\{ \forall x : [\exists y : \lnot B(x,y)] \}\tag{2B}$$

Using the given Example :
$$\{ \forall x : [\exists y : xy\not=0] \}\tag{2C}$$

Domain Integers will make that false : for all integers $x$ , we can not get some $y$ which makes $xy\not=0$ , because we might have $x=0$ then there is no $y$ to ensure $0y\not=0$

Domain Irrational numbers will make that true : for all Irrational numbers $x$ , we can get some $y=1/x$ which makes $xy=1$ , thus we can ensure $xy\not=0$

SUMMARY :

The given Statement is (1)
The Negation is (2A) when we consider only the outer quantifier
The Negation is (2B) when we consider the outer quantifier with the inner quantifier too
The Negation is (2C) when we consider the outer quantifier with the inner quantified too , while considering the Example $B$ to change $=$ to $\not =$, though that is not very general.

ADDENDUM :

The "Standard" laws are well known. Check here :
https://math.libretexts.org/Courses/SUNY_Schenectady_County_Community_College/Discrete_Structures/02%3A_Logical_Reasoning/2.04%3A_Quantifiers_and_Negations

It is just :

$$\lnot \forall x : X(x) \equiv \exists x : \lnot X(x)$$ $$\lnot \exists x : X(x) \equiv \forall x : \lnot X(x)$$

Answer 3

In my opinion the statement

there does not exist a $x$ for all $y$, $B(x,y)$

is a bit ambiguous, and I tend to interpret it differently to the other answers. I tend to read the following statement, without the negation,

there exists an $x$ for all $y$ [such that] $B(x,y)$

as $$\forall y\ \exists x\ B(x,y),$$ similarly to how you say "there is a lid for every pot", which in my understanding means that for every pot there exists a lid (which fits it) and not the other way around. I read the original statement as its negation, i.e. $$\lnot(\forall y(\exists x(B(x,y))))=\exists y(\forall x(\lnot B(x,y))).$$

On the other hand, if this is an exercise question, the choice of variables $x,y$ and the order in which they appear in $B(x,y)$ suggest that $x$ comes first in the formula and then $y$ ;) So @ryang's and @Prems' answers are probably the desired solutions to the question in this case.