1.6 Nested Quantiﬁersbtravers.weebly.com/uploads/6/7/2/9/6729909/nested... · When we talk of nested quantiﬁers, we mean statements like 8x 9y (x

§ 1.6 Nested Quantifiers

The Beginning

When we talk of nested quantifiers, we mean statements like

∀ x ∃ y (x < y)

This is the same thing as ∀ x Q(x), where Q(x) is ∃ y P(x, y), whereP(x, y) is x > y. This is where the term ‘nested’ comes from.

To understand the statements involving multiple quantifiers, we needto make sure we understand what the quantifiers and predicates thatwe see mean.

The Beginning


∀ x ∃ y (x < y)



The Beginning


∀ x ∃ y (x < y)



Examples

ExampleAssume that the domain of discourse for the variables x and y consistof all real numbers (R×R).

The statement∀ x ∀ y (x + y = y + x)

says that x + y = y + x for all real numbers x and y. This is thecommutative law for the addition of real numbers.

We could have written this in terms of the predicate P(x, y) whereP(x, y) is x + y = y + x.

Likewise, the statement

∀ x ∃ y (x + y = 0)

says that for every real number x there is a real number y such thatx + y = 0. This is the additive inverse property.

Examples

ExampleAssume that the domain of discourse for the variables x and y consistof all real numbers (R×R).The statement

∀ x ∀ y (x + y = y + x)




∀ x ∃ y (x + y = 0)


Examples


∀ x ∀ y (x + y = y + x)




∀ x ∃ y (x + y = 0)


Examples


∀ x ∀ y (x + y = y + x)




∀ x ∃ y (x + y = 0)


Examples


∀ x ∀ y (x + y = y + x)




∀ x ∃ y (x + y = 0)


Examples (cont.)

ExampleFinally, the statement

∀ x ∀ y ∀ z (x + (y + z) = (x + y) + z)

is the associative law for addition of real numbers.

ExampleAssume the domain of discourse isR×R. Translate the followinginto English:

∀ x ∀ y ((x > 0) ∧ (y < 0)→ (xy < 0))

The statement says that for every real number x and every real numbery, if x > 0 and y < 0, then xy < 0. That is, the product of a positivereal number and a negative real number is a negative real number.

Examples (cont.)


∀ x ∀ y ∀ z (x + (y + z) = (x + y) + z)



∀ x ∀ y ((x > 0) ∧ (y < 0)→ (xy < 0))


Examples (cont.)


∀ x ∀ y ∀ z (x + (y + z) = (x + y) + z)



∀ x ∀ y ((x > 0) ∧ (y < 0)→ (xy < 0))


Does Order Matter?

Example

Assume the domain of discourse isR×R. Let P(x, y) be thestatement ‘x + y = y + x’. What are the truth values for thequantification ∀ x ∀ y P(x, y) and ∀ y ∀ x P(x, y)?

Because P(x, y) is true for all real numbers x and all real numbers y,the statement ∀ x ∀ y P(x, y) is true.

Notice that ∀ y ∀ x P(x, y) states that for all real numbers y and all realnumbers x, x + y = y + x. So both statements have the same meaning.

Does Order Matter?

Example




Does Order Matter?

Example




Does Order Matter

Example

Assume the domain of discourse isR×R. Let Q(x, y) be thestatement x + y = 0. What are the truth values for the quantification∀ x ∃ y Q(x, y) and for ∃ y ∀ x Q(x, y)?

We have already seen ∀ x ∃ y (Q(x, y)) and we know this is a truestatement.

What does the statement ∃ y ∀ x (Q(x, y)) say?

The statement says ‘There is a number y such that for every realnumber x, Q(x, y).

No matter what value of y is chosen, there is only one value x thatmakes x + y = 0 true, so the statement ∃ y ∀ x Q(x, y) is false.

Does Order Matter

Example






Does Order Matter

Example






Does Order Matter

Example






Does Order Matter

Example






Does Order Matter

Conclusion? Do you think order matters with nested quantifiers?

What we have is

∀ x ∀ y P(x, y) ≡ ∀ y ∀ x P(x, y)

∃ y ∀ x Q(x, y) 6≡ ∀ x ∃ y Q(x, y)

This makes it seem that order matters when we use differentquantifiers.

Does Order Matter


What we have is

∀ x ∀ y P(x, y) ≡ ∀ y ∀ x P(x, y)

∃ y ∀ x Q(x, y) 6≡ ∀ x ∃ y Q(x, y)


Does Order Matter


What we have is

∀ x ∀ y P(x, y) ≡ ∀ y ∀ x P(x, y)

∃ y ∀ x Q(x, y) 6≡ ∀ x ∃ y Q(x, y)


Quantification of Two Variables

Quantifications of Two VariablesStatement When True? When False?∀ x ∀ y P(x, y) P(x, y) is true for There is a pair for∀ y ∀ x P(x, y) all pairs x, y which P(x, y) is false

∀ x ∃ y P(x, y) For every x there is a y There is an x such thatfor which P(x, y) is true P(x, y) is false for every y

∃ x ∀ y P(x, y) There is an x for which For every x there is a y forP(x, y) is true for every y which P(x, y) is false

∃ x ∃ y P(x, y) There is a pair x, y for P(x, y) is false for every∃ y ∃ x P(x, y) which P(x, y) is true pair x, y


Quantifications of Two VariablesStatement When True? When False?∀ x ∀ y P(x, y) P(x, y) is true for There is a pair for∀ y ∀ x P(x, y) all pairs x, y which P(x, y) is false∀ x ∃ y P(x, y) For every x there is a y There is an x such that

for which P(x, y) is true P(x, y) is false for every y

∃ x ∀ y P(x, y) There is an x for which For every x there is a y forP(x, y) is true for every y which P(x, y) is false




for which P(x, y) is true P(x, y) is false for every y∃ x ∀ y P(x, y) There is an x for which For every x there is a y for

P(x, y) is true for every y which P(x, y) is false




for which P(x, y) is true P(x, y) is false for every y∃ x ∀ y P(x, y) There is an x for which For every x there is a y for

P(x, y) is true for every y which P(x, y) is false∃ x ∃ y P(x, y) There is a pair x, y for P(x, y) is false for every∃ y ∃ x P(x, y) which P(x, y) is true pair x, y

An Example

Example

Let Q(x, y, z) be the statement x + y = z. What are the truth values ofthe statements ∀ x ∀ y ∃ z Q(x, y, z) and ∃ z ∀ x ∀ y Q(x, y, z), wherethe domain of discourse isR3?

What does the statement ∀ x ∀ y ∃ z Q(x, y, z) mean?

‘For all real numbers x and for all real numbers y there is a realnumber z such that it is true that x + y = z’

Is the quantification a true statement? YES

An Example

Example





An Example

Example





An Example

Example




Is the quantification a true statement?

YES

An Example

Example





An Example

What about if we rearrange and look at ∃ z ∀ x ∀ y Q(x, y, z). What isthe meaning here?

‘There is a real number z such that for all real numbers x and for allreal numbers y it is true that x + y = z’.

Is this true or false?

This is false since there is no value for z that satisfies the equationx + y = z for all values of x and y.

An Example





An Example





An Example





Translating Mathematical Statements

ExampleTranslate the statement ‘The sum of two positive integers is alwayspositive’ into a logical expression.

First, what is the domain of discourse?

The domain of discourse is Z × Z .

∀ x ∀ y ((x > 0) ∧ (y > 0)→ (x + y > 0))

Alternately, we can write this as

∀ x ∀ y (x + y > 0)

where the domain of discourse consists of all positive integers.





∀ x ∀ y ((x > 0) ∧ (y > 0)→ (x + y > 0))


∀ x ∀ y (x + y > 0)






∀ x ∀ y ((x > 0) ∧ (y > 0)→ (x + y > 0))


∀ x ∀ y (x + y > 0)






∀ x ∀ y ((x > 0) ∧ (y > 0)→ (x + y > 0))


∀ x ∀ y (x + y > 0)






∀ x ∀ y ((x > 0) ∧ (y > 0)→ (x + y > 0))


∀ x ∀ y (x + y > 0)


Another Example

ExampleTranslate the statement ‘Every real number except zero has amultiplicative inverse’.

∀ x (x 6= 0)→ ∃ y (xy = 1)

Another Example

ExampleTranslate the statement ‘Every real number except zero has amultiplicative inverse’.

∀ x (x 6= 0)→ ∃ y (xy = 1)

And Another

ExampleExpress the statement using predicates and quantifiers with a domainof discourse consisting of all people.‘If a person is female and is a parent, then this person is someone’smother’.

First we define the propositional functions. How many do we needhere?

Let F(x) represent ‘x is a female‘Let P(x) represent x is a parentLet M(x, y) represent x is the mother of y

So, our statement would be ...

∀ x ((F(x) ∧ P(x))→ ∃ y M(x, y)

And Another





∀ x ((F(x) ∧ P(x))→ ∃ y M(x, y)

And Another



Let F(x) represent

‘x is a female‘Let P(x) represent x is a parentLet M(x, y) represent x is the mother of y


∀ x ((F(x) ∧ P(x))→ ∃ y M(x, y)

And Another



Let F(x) represent ‘x is a female‘

Let P(x) represent x is a parentLet M(x, y) represent x is the mother of y


∀ x ((F(x) ∧ P(x))→ ∃ y M(x, y)

And Another



Let F(x) represent ‘x is a female‘Let P(x) represent

x is a parentLet M(x, y) represent x is the mother of y


∀ x ((F(x) ∧ P(x))→ ∃ y M(x, y)

And Another



Let F(x) represent ‘x is a female‘Let P(x) represent x is a parent

Let M(x, y) represent x is the mother of y


∀ x ((F(x) ∧ P(x))→ ∃ y M(x, y)

And Another



Let F(x) represent ‘x is a female‘Let P(x) represent x is a parentLet M(x, y) represent

x is the mother of y


∀ x ((F(x) ∧ P(x))→ ∃ y M(x, y)

And Another





∀ x ((F(x) ∧ P(x))→ ∃ y M(x, y)

Still Another Example

ExampleExpress the statement ‘Everyone has exactly one best friend’ usingpredicates and quantifiers.

The problem isn’t stating that someone has a best friend, it is statingthat they have exactly one best friend. Thoughts?

We can think of this as ‘For every person x, x has exactly one bestfriend’. To say that x has exactly one best friend implies that there is aperson y who has x as their best friend. Further, for every person zwho is not y then z is not the best friend of y.

Let B(x, y) be the statement ‘x is the best friend of y’.

∀ x ∃ y B(x, y) ∧ ∀ z ((z 6= y)→ ¬B(x, z))






∀ x ∃ y B(x, y) ∧ ∀ z ((z 6= y)→ ¬B(x, z))






∀ x ∃ y B(x, y) ∧ ∀ z ((z 6= y)→ ¬B(x, z))






∀ x ∃ y B(x, y) ∧ ∀ z ((z 6= y)→ ¬B(x, z))






∀ x ∃ y B(x, y) ∧ ∀ z ((z 6= y)→ ¬B(x, z))

Still More Examples

ExampleUse quantifiers to express ‘There is a woman who has taken a flighton every airline in the world’.

Propositional functions?

Let P(w, f ) be ‘w has taken f ’Q(f , a) be ‘f is a flight on a’

What is the domain of discourse?

The domain of discourse would be all flights f , all airlines a and allwomen w.

And the statement.

∃ w ∀ a ∃ f (P(w, f ) ∧ Q(f , a))

Still More Examples



Let P(w, f ) be

‘w has taken f ’Q(f , a) be ‘f is a flight on a’



And the statement.

∃ w ∀ a ∃ f (P(w, f ) ∧ Q(f , a))

Still More Examples



Let P(w, f ) be ‘w has taken f ’

Q(f , a) be ‘f is a flight on a’



And the statement.

∃ w ∀ a ∃ f (P(w, f ) ∧ Q(f , a))

Still More Examples



Let P(w, f ) be ‘w has taken f ’Q(f , a) be

‘f is a flight on a’



And the statement.

∃ w ∀ a ∃ f (P(w, f ) ∧ Q(f , a))

Still More Examples






And the statement.

∃ w ∀ a ∃ f (P(w, f ) ∧ Q(f , a))

Still More Examples






And the statement.

∃ w ∀ a ∃ f (P(w, f ) ∧ Q(f , a))

Still More Examples






And the statement.

∃ w ∀ a ∃ f (P(w, f ) ∧ Q(f , a))

Still More Examples






And the statement.

∃ w ∀ a ∃ f (P(w, f ) ∧ Q(f , a))

Still More Examples






And the statement.

∃ w ∀ a ∃ f (P(w, f ) ∧ Q(f , a))

Remember Calculus?

ExampleExpress the definition of a limit using quantifiers.

Does anyone remember/know the definition of the statement

limx→a

f (x) = L

For every real number ε > 0, there exists a real number δ > 0 suchthat |f (x)− L| < ε whenever 0 < |x− a| < δ.

So how can we represent this with quantifiers?

∀ ε ∃ δ (0 < |x− a| < δ → |f (x)− L| < ε)

Remember Calculus?



limx→a

f (x) = L



∀ ε ∃ δ (0 < |x− a| < δ → |f (x)− L| < ε)

Remember Calculus?



limx→a

f (x) = L



∀ ε ∃ δ (0 < |x− a| < δ → |f (x)− L| < ε)

Remember Calculus?



limx→a

f (x) = L



∀ ε ∃ δ (0 < |x− a| < δ → |f (x)− L| < ε)

Remember Calculus?



limx→a

f (x) = L



∀ ε ∃ δ (0 < |x− a| < δ → |f (x)− L| < ε)

Translating From Nested Quantifiers into English

ExampleTranslate the statement

∀ x (C(x) ∨ ∃ y (C(y) ∧ F(x, y)))

where C(x) is ‘x has a laptop’, F(x, y) is ‘x and y are friends’ and thedomain of discourse is all students at SSU.

For every student x at SSU, x has a laptop or there is a student y atSSU such that y has a laptop and x and y are friends.



∀ x (C(x) ∨ ∃ y (C(y) ∧ F(x, y)))

where C(x) is ‘x has a laptop’, F(x, y) is ‘x and y are friends’ and thedomain of discourse is all students at SSU.

For every student x at SSU, x has a laptop or there is a student y atSSU such that y has a laptop and x and y are friends.



∃ x ∀ y ∀ z ((F(x, y) ∧ F(x, z) ∧ (y 6= z))→ ¬F(y, z))

into English, where F(a, b) means a and b are friends and the domainof discourse for x, y and z consist of all SSU students.

For some SSU student x, for all SSU student y and for all SSU studentz, if x and y are friends, and if x and z are friends and if y and z are notthe same person, then y and z are not friends. In other words, there areno students whose friends are also friends.



∃ x ∀ y ∀ z ((F(x, y) ∧ F(x, z) ∧ (y 6= z))→ ¬F(y, z))

into English, where F(a, b) means a and b are friends and the domainof discourse for x, y and z consist of all SSU students.

For some SSU student x, for all SSU student y and for all SSU studentz, if x and y are friends, and if x and z are friends and if y and z are notthe same person, then y and z are not friends. In other words, there areno students whose friends are also friends.

Negating Quantifiers

ExampleNegate the statement

∀ x ∃ y (xy = 1)

There are three different ways we can express this:

1 ∃ x ¬∃ y (xy = 1)2 ∃ x ∀ y ¬(xy = 1)3 ∃ x ∀ y (xy 6= 1)



∀ x ∃ y (xy = 1)


1 ∃ x ¬∃ y (xy = 1)2 ∃ x ∀ y ¬(xy = 1)3 ∃ x ∀ y (xy 6= 1)



∀ x ∃ y (xy = 1)


1 ∃ x ¬∃ y (xy = 1)

2 ∃ x ∀ y ¬(xy = 1)3 ∃ x ∀ y (xy 6= 1)



∀ x ∃ y (xy = 1)


1 ∃ x ¬∃ y (xy = 1)2 ∃ x ∀ y ¬(xy = 1)

3 ∃ x ∀ y (xy 6= 1)



∀ x ∃ y (xy = 1)


1 ∃ x ¬∃ y (xy = 1)2 ∃ x ∀ y ¬(xy = 1)3 ∃ x ∀ y (xy 6= 1)

One Final (Long) Example

Example

Let F(x, y) be the statement ‘x can fool y’, where the domain ofdiscourse consists of all people in the world. Use quantifiers toexpress each of the following statements:

1 Everybody can fool Jack

∀ x F(x, Jack)2 Owen can fool everybody∀ x F(Owen, x)

3 Everybody can fool somebody∀ x ∃ y F(x, y)

4 There is no one who can fool everybody∀ x ∃ y ¬F(x, y)


Example


1 Everybody can fool Jack∀ x F(x, Jack)

2 Owen can fool everybody∀ x F(Owen, x)




Example



2 Owen can fool everybody

∀ x F(Owen, x)3 Everybody can fool somebody∀ x ∃ y F(x, y)



Example







Example




3 Everybody can fool somebody

∀ x ∃ y F(x, y)4 There is no one who can fool everybody∀ x ∃ y ¬F(x, y)


Example







Example





4 There is no one who can fool everybody

∀ x ∃ y ¬F(x, y)


Example







Example


1 Everybody can be fooled by somebody

∀ x ∃ y F(y, x)2 No one can fool both Jack and Owen∀ x (F(x,Owen)⊕ F(x, Jack))

3 Jack can fool exactly two people∃ x ∃ y (x 6= y ∧ F(Jack, x) ∧ F(Jack, y) ∧ ∀ z (F(Jack, z)→(z = x ∨ z = y)))

4 No one can fool themself∀ x ¬F(x, x)


Example


1 Everybody can be fooled by somebody∀ x ∃ y F(y, x)

2 No one can fool both Jack and Owen∀ x (F(x,Owen)⊕ F(x, Jack))




Example



2 No one can fool both Jack and Owen

∀ x (F(x,Owen)⊕ F(x, Jack))3 Jack can fool exactly two people∃ x ∃ y (x 6= y ∧ F(Jack, x) ∧ F(Jack, y) ∧ ∀ z (F(Jack, z)→(z = x ∨ z = y)))



Example







Example




3 Jack can fool exactly two people

∃ x ∃ y (x 6= y ∧ F(Jack, x) ∧ F(Jack, y) ∧ ∀ z (F(Jack, z)→(z = x ∨ z = y)))



Example







Example





4 No one can fool themself

∀ x ¬F(x, x)


Example






Documents

1.6 Nested Quantiﬁersbtravers.weebly.com/uploads/6/7/2/9/6729909/nested... · When we talk of nested quantiﬁers, we mean statements like 8x 9y (x