Methods of Proof for Quantifiers

1
Methods of Proof for Quantifiers
Language, Proof and Logic
Chapter 12
2
Valid quantifier steps
12.1
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. ?xCube(x)?Large(x) 2. ?xLarge(x)?LeftOf(x,
b) 3. Cube(d) 4. ?xLarge(x)?LeftOf(x,b)
Let us think about whether there is any
similarity with ?-elim and ?-intro.
3
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.
1. ?xCube(x)?Large(x) 2. ?xLarge(x)?LeftOf(x,
b) 3. ?xCube(x) 4. ?xLarge(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 db, we would have
been able to prove ?xLeftOf(x,x)!
Let us think about whether there is any
similarity with ?-elim.
4
The method of general conditional proof
12.3.a
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).
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.

• 1. ?xLarge(x)
• 2. ?xLarge(x)?SameRow(x,b)
• 3. ?xSameRow(x,b)

Important The arbitrary object
1. Cube(b) d indeed has to be arbitrary. Things
2. ?xCube(x)?Large(x) will go wrong
if you select db here
3.
?xLarge(x)
Let us think about whether there is any
similarity with ?-intro.
5
The method of general conditional proof
12.3.b
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
?xP(x)?Q(x).

• 1. ?xCube(x)?SameRow(x,b)
• 2. ?xSameRow(x,b)?Small(x)
• 3. ?xCube(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.
6
Proofs involving mixed quantifiers
12.4.a
1. ?yGirl(y) ? ?x(Boy(x) ? Likes(x,y)) 2.
?xBoy(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. ?xBoy(x) ? ?y(Girl(y) ? Likes(x,y)) 2.
?yGirl(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.
7
Proofs involving mixed quantifiers
12.4.b

• 
  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 ?xP(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.

8
Proofs involving mixed quantifiers
12.4.c
Euclids Theorem ?x?yy?x ? Prime(y)
Proof. Consider an arbitrary natural number n.
Our goal is to show that ?yy?n ? Prime(y),
from which Euclids theorem follows by
universal generalization. Let k be the product
of all the prime numbers less than n. Thus each
prime with ltn divides k without remainder. Now
let mk1. Each prime less then 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?yy?x ?
Prime(y).
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Proofs involving mixed quantifiers
12.4.d
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.