methods of proof for quantifiers l.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
Methods of Proof for Quantifiers PowerPoint Presentation
Download Presentation
Methods of Proof for Quantifiers

Loading in 2 Seconds...

play fullscreen
1 / 9

Methods of Proof for Quantifiers - PowerPoint PPT Presentation


  • 302 Views
  • Uploaded on

Language, Proof and Logic. Methods of Proof for Quantifiers. Chapter 12. 12.1. Valid quantifier steps. Universal elimination ( instantiation ): From  xP(x) infer P(c). where c is the name of some object of the domain of discourse. Existential introduction ( generalization ):

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

Methods of Proof for Quantifiers


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
valid quantifier steps

12.1

Valid quantifier steps

Universal elimination (instantiation):

From xP(x) infer P(c)

where c is the name

of some object of the

domain of discourse

Existential introduction (generalization):

From P(c) infer xP(x)

3 says that d is a cube. And

1 says that all cubes are large.

Thus, d is large. But 2 says

that every large object is to

the left of b. So, d is to the

left of b. To summarize,

d is large and is to the left of b.

Thus, there is a large object to

the left of b.

1. x[Cube(x)Large(x)]

2. x[Large(x)LeftOf(x,b)]

3. Cube(d)

4. x[Large(x)LeftOf(x,b)]

Let us think about whether there is any similarity with -elim and-intro.

the method of existential instantiation

12.2

Existential instantiation (elimination): Once you have proven xP(x)

(or have it as a premise), you can select a “neutral” (not used elsewhere)

name d and use P(d) as a valid assumption.

The method of existential instantiation

1. x[Cube(x)Large(x)]

2. x[Large(x)LeftOf(x,b)]

3. xCube(x)

4. x[Large(x)LeftOf(x,b)]

3 says that there is a cube. Let d

be such a cube, i.e. assume

Cube(d) (is true).

1 says that all cubes are large.

Thus, d is large. But 2 says

that every large object is to

the left of b. So, d is to the

left of b. To summarize,

d is large and is to the left of b.

Thus, there is a large object to

the left of b.

Important: If we had selected

d=b, we would have been able to

“prove” xLeftOf(x,x)!

Let us think about whether there is any similarity with-elim.

the method of general conditional proof

12.3.a

The method of general conditional proof

Universal generalization (introduction): Once you have proven P(d) for

some “neutral” (not used elsewhere) name d (denoting a “totally

arbitrary” object), you can conclude xP(x).

  • 1. xLarge(x)
  • 2. x[Large(x)SameRow(x,b)]
  • 3. xSameRow(x,b)

Consider any object d. By 1, d is

large. But, by 2, every large

object is in the same row as b. So,

d is in the same row as b. As d

was arbitrary, we conclude that

every object is in the same row

as b.

Important: The “arbitrary” object 1. Cube(b)

d indeed has to be arbitrary. Things 2. x[Cube(x)Large(x)]

will go wrong if you select d=b here

3. xLarge(x)

Let us think about whether there is any similarity with-intro.

the method of general conditional proof5

12.3.b

The method of general conditional proof

General conditional proof: Once you have proven Q(d) from the

assumption P(d) for some “neutral” (not used elsewhere) name d

(denoting a “totally arbitrary” object), you can conclude x[P(x)Q(x)].

  • 1. x[Cube(x)SameRow(x,b)]
  • 2. x[SameRow(x,b)Small(x)]
  • 3. x[Cube(x)Small(x)]

Consider any object d, and assume d

is a cube. 1 says that every cube is in

the same row as b. So, d is in the same

row as b. But, by 2, everything in the

same row as b is small. So, d is small.

As d was arbitrary, we conclude that

every cube is small.

Let us think about why universal generalization in fact makes this rule redundant.

proofs involving mixed quantifiers

12.4.a

Proofs involving mixed quantifiers

1.y[Girl(y)  x(Boy(x)  Likes(x,y))]

2.x[Boy(x) y(Girl(y)  Likes(x,y))]

Consider an arbitrary boy d. By 1, there is a girl who is liked by every

boy. Let c be such a girl. So, d likes c. That is, d likes some girl. As

d was arbitrary, we conclude that every boy likes some girl.

1.x[Boy(x) y(Girl(y)  Likes(x,y))]

2. y[Girl(y)  x(Boy(x)  Likes(x,y))]

Pseudo-proof: Consider an arbitrary boy d. By 1, d likes some girl. Let

c be such a girl. Thus, d likes c. Since d was arbitrary, we conclude that

every boy likes c. So, there is a girl (specifically, c) who is liked

by every boy.

proofs involving mixed quantifiers7

12.4.b

Proofs involving mixed quantifiers
  • REMEMBER
  • Let P(x), Q(x) be wffs.
  • Existential Instantiation: If you have proven xP(x) then you may
  • choose a new constant symbol c to stand for any object satisfying
  • P(x) and so you may assume P(c).
  • 2. General Conditional Proof: If you want to prove x[P(x)Q(x)]
  • then you may choose a new constant symbol c, assume P(c), and
  • prove Q(c), making sure that Q does not contain any names
  • introduced by existential instantiation after the assumption of P(c).
  • 3. Universal Generalization: If you want to prove xQ(x) then you
  • may choose a new constant symbol c and prove Q(c), making sure
  • that Q does not contain any names introduced by existential
  • instantiation after the introduction of c.
proofs involving mixed quantifiers8

12.4.c

Proofs involving mixed quantifiers

Euclid’s Theorem: xy[yx  Prime(y)]

Proof. Consider an arbitrary natural number n. Our goal is to show that

y[yn  Prime(y)],from which Euclid’s theorem follows by universal

generalization.

Let k be the product of all the prime numbers less than n. Thus each

prime with <n divides k without remainder. Now let m=k+1. Each

prime less than n divides m with remainder 1. But we know that m can

be factored into primes. Let p be one of those primes. Clearly, by the

earlier observation, pn. Hence, by existential generalization, there

is a prime (specifically, p) greater or equal to n.

As n was arbitrary, we conclude that xy[yx  Prime(y)].

proofs involving mixed quantifiers9

12.4.d

Proofs involving mixed quantifiers

The Barber Paradox:

xy [Shave(x,y) Shave(y,y)]

The domain of discourse is the set of all men in a small village.

Proof. Assume, for a contradiction, that

1. xy [Shave(x,y)  Shave(y,y)]

Let b be a man (barber) such that

2.y [Shave(b,y)  Shave(y,y)]

is true. By universal instantiation from 2,

3. Shave(b,b)  Shave(b,b).

But this is (indeed) a contradiction.