Methods of Proof for Quantifiers

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# Methods of Proof for Quantifiers - PowerPoint PPT Presentation

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 ):

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## Methods of Proof for Quantifiers

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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.

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.

d=b, we would have been able to

“prove” xLeftOf(x,x)!

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

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.

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.

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.

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.

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)].

12.4.d

Proofs involving mixed quantifiers

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.