Download Section 1.5 Proofs in Predicate Logic

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Section 1.5
Proofs in Predicate Logic
Section 1.5:
1.5: Proofs in Predicate Logic
Purpose of Section:
Section The theorems in Section 1.4 included quantifiers although
the theorems were not stated explicitly in the language of predicate logic. In
this section we state theorems in predicate logic language and show different
theorems in predicate logic relate to one another.
Introduction
Most theorems in mathematics include quantifiers like “for all”
and “there exists” although they are often understood and not stated
explicitly. For example the theorem “the square of an even number n is
even” does not explicitly say that n is an integer. A more formal statement of
the theorem might use the language of predicate logic and
(
)
write ( ∀x ∈ ) x even ⇒ x 2 even . In this section analyze proofs of theorems
stated in the formal language of predicate logic.
Theorems Involving Universal and Existential Quantifiers
Universal Sentences: To prove a theorem of the form
( ∀x ∈U ) P( x)
which means “for all elements x in a given universe U , the proposition P ( x)
is true” we select an arbitrary x ∈ U from the universe, and then prove the
assertion P ( x) is true.
Theorem 1 (Universal Direct Proof) Show that all integers divisible by 6 are
even.
Proof: In the language of predicate logic, we write
( ∀x ∈ ) ( 6 x ⇒ x is even )
where = {0, ± 1, ± 2,...} is the universe of integers. Letting x be an arbitrary
integer, we assume x is divisible by 6, which means there exists an integer
m which satisfies x = 6m . Rewriting this as x = 2 ( 3m ) = 2k , and since k = 3m
is also an integer we know x is an even integer.
▌
Proofs involving the existential quantifier ∃ are easier. We only need to find
one element in the universe that satisfies the given proposition.
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Section 1.5
Proofs in Predicate Logic
Existential Sentences: To prove a theorem of the form
( ∃x ∈U ) P( x)
which means “there exists an element x in a given universe U that satisfies
the proposition P ( x) ” the strategy is to find one (or more) elements x ∈ U
that satisfy P ( x) .
Theorem 2
number.
(Proof
(Proof by Demonstration)
Demonstration)
Show there exists an even prime
Proof: In the language of predicate logic, we would write
( ∃x ∈ )( x is even ∧ x is prime )
where is the set of natural numbers.
both even and prime.
The proof is trivial. The number 2 is
▌
Some theorems contain both universal and existential quantifiers.
Theorem 3 (Proof by Demonstration) There exists irrational numbers a, b
such that a b is rational.
Proof
Recall from Section 1.4 that
2
2 .
2 is irrational. Now consider the number
Either the number is rational or irrational. If the number is rational,
we have proven the theorem with a = b = 2 .
consider
If
2
2
is irrational, then
2
a= 2 ,b= 2
and observe  2

2



2
=
( 2)
2
= 2 . In this case we have proven the theorem
2
with a = 2 , b = 2 .
▌
Theorem 4: (Existential Direct Proof)
In the universe of real numbers show
( ∀x ∈ )( ∃y ∈ )( x + y = 1) .
Proof: In plain English this theorem reads “for every real number x there
exists a real number y which satisfies x + y = 1 . Letting x be an arbitrary
real number, it is clear that the real number y = 1 − x satisfies
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Section 1.5
Proofs in Predicate Logic
x + y = x + (1 − x ) = 1 .
Note that interchanging the quantifiers to
▌
( ∃x )( ∀y )( x + y = 1)
makes the
theorem false; i.e. there does not exist a real number x such that for all real
numbers y the equation x + y = 1 holds. As a counterexample select x = 2 ,
then clearly 2 + y = 1 is not true for all y ( y = 3 fails).
Proofs by Contradiction for Universal and Existential Quantifiers
Proof of ( ∀x ) P ( x) by Contradiction:
To prove the theorem ( ∀x ) P ( x) which says “for all x , P ( x) is true” by
contradiction, assume the contrary; i.e. ∼ ( ∀x ) P ( x) which is equivalent to
( ∃x ) ∼ P( x) , which says “there exists an
x such that P ( x) is not true”. One
then continues the proof until arriving at a contradiction.
Theorem 5:
5: (Proof by Contradiction)
Contradiction)
Show if m, n are integers, then 14m + 21n ≠ 1 .
Proof: In the language of predicate logic this theorem becomes
( ∀m, n ∈ )(14m + 21n ≠ 1)
Since it is easier to work with equations rather than inequalities, we assume
the contrary, which says
( ∃m, n ∈ )(14m + 21n = 1)
But if this equation says the natural number 7 divides the left side of the
equation but not the right, which cannot be true. Hence, the denial is false so
the theorem is true.
▌
Note: Some theorems seem obvious but are incredible difficult to prove. Onec
such theorem is the Jordan Curve Theorem, which states that every closed
curve in the plane divides the plane into two an “inside” and an “outside. The
theorem was posed by the French and proven by the American geometer
Oswald Veblen in 1905. In 2005 an international team of mathematicians using a
formal computer checking system called Mizar generated a 6,500 line proof of the
theorem..
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Section 1.5
Theorem 6:
6:
2
(Proof by Contradiction)
Proofs in Predicate Logic
If x, y are positive integers, then
2
x − y ≠ 1.
Proof
In the language of predicate logic, the theorem is
( ∀x ∈ )( ∀y ∈ ) ( x 2 − y 2 ≠ 1)
Assuming the contrary, we have
( ∃x ∈ )( ∃y ∈ ) ( x 2 − y 2 = 1)
and factoring yields
x 2 − y 2 = ( x − y )( x + y ) = 1 .
Now,
since
are positive integers it follows that both factors
x − y and x + y are either +1 or both factors are −1. In the first case, if both
factors are +1 , then adding the factors yields x = 1, y = 0 contradicting the
assumption that x, y are both positive. In the second case, when both factors
are −1 , if the factors are added one gets x = −1, y = 0 , again contradicting the
fact both x, y are positive. Hence the denial is false and thus the theorem is
true. ▌
x, y
Intuitionism: In the philosophy of mathematics there is a school of thought,
called Intuitionism, with past proponents like the Dutch mathematician L.E.J.
Brouwer (1881-1961) and German mathematician Leopold Kronecker (18231891). Intuitionists feel that mathematics should constructions based on the
“intuitively given” natural numbers, 1,2,3,…
and not on logical axioms.
Kronecker once said, “God made the integers, all else is the work of man.” In
the late 1800s and early 1900s several intuitionists felt that the new theories
which were sweeping mathematics at the time, like Cantor’s transfinite
arithmetic, infinite sets, imaginary numbers, proof by contradictions, noneuclidean geometries and so on were taking mathematics down the road to
mysticism and fantasy. Even the great French mathematician Henri Poincare
(1854-1912) felt that Cantor’s theory of infinite sets and transfinite arithmetic
should be excluded from mathematics.
Today the intuitionist school of
mathematics is not held in favor among many mathematicians.
It is also possible to deny an existential quantifier.
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Section 1.5
Proofs in Predicate Logic
Proof of ( ∃x ) P ( x) by Contradiction:
To prove “there exists an x such that P ( x) is true” assume the theorem
is not true: i.e. ∼ ( ∃x ) P ( x) ≡ ( ∀x ) ∼ P ( x) which states “for all x the assertion
P ( x) is not true.
You then continue the proof until you arrive at a
contradiction of some kind.
Unique Existential Quantification
Proving Unique Existential Theorems: A theorem of the form
( ∃! x ∈U ) P( x)
states “there exists a unique element x such that P ( x) is true.”
To prove
this theorem we must show P ( x ) for exactly one element of the universe. A
common strategy is to show first that some element x satisfies P ( x) , then
show that if two elements y, z ∈ U satisfy the assertion, then in fact they are
the same; i.e. y = z. In predicate logic language, we must show
▪
▪
( ∃x ∈ U ) P ( x )
( ∀y ∈U )( ∀z ∈U ) [ P( y ) ∧ P( z ) ⇒ y = x ]
Theorem 7 Show that the differential equation with initial condition
dy
+ y = 1, y (0) = 0
dt
has exactly one solution1.
Solution:
Solution: It can easily be verified that y = 1 − e −t satisfies both the differential
equation and the initial condition. Hence, we have shown existence by simply
demonstrating a solution.2 To show uniqueness suppose y1 (t ) and y2 (t ) both
satisfy the differential equation and initial condition, which means
dy1
+ y1 = 1, y1 (0) = 0
dt
1
dy2
+ y2 = 1, y2 (0) = 0
dt
For readers unfamiliar with differential equations, the goal of this example is to show how uniqueness of
solutions is obtained. Simply take facts presented at face value.
2
There are standard methods for finding the solution of this problem.
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Section 1.5
Proofs in Predicate Logic
Subtracting the respective above equations, we find
 dy1
  dy
 d
+ y1  −  2 + y2  = ( y1 − y2 ) + ( y1 − y2 ) = 0

 dt
  dt
 dt
y1 (0) − y2 ( 0 ) = 0
If we now let Y = y1 − y2 the above equations take the form
d
Y (t ) + Y (t ) = 0
dt
Y ( 0) = 0
But it can also be shown that the only function that simultaneously satisfies
these two equations is the zero solution Y (t ) ≡ 0 . Hence, we have y1 − y2 = 0
or y1 = y2 which proves uniqueness of the problem.
▌
Theorem 8 (Uniqueness
(Uniqueness : Diophantine Equation)
(
Prove ( ∃!m ∈ )( ∃!∈ ) m 2 − n 2 = 12
)
In other words, there exists unique natural numbers m and n that satisfy the
equation m 2 − n 2 − 12 . (An equation which restricts the variables to integers
is called a Diophantine equation.)
Proof: We factor the difference between the two squares as
m 2 − n 2 = ( m + n )( m − n ) = 12
and make the observation that the difference between the factors
m + n and m − n is 2n , which tells us that the two factors of 12 are both odd
or both even. Hence we must have
m+n = 6
m−n = 2
and solving for m, n gives m = 4, n = 2 . But this is the only solution since it
represents the only way to factor 12 into two natural numbers both of which
are even.
▌
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Section 1.5
Proofs in Predicate Logic
Problems
1.
(True or False?) Which of the following are true?
a)
i)
( ∀x ∈ ℜ ) ( x 2 + x + 1 > 0 )
( ∀x ∈ ℜ )  x 2 > 0 ∨ x 2 < 0 
( ∀x ∈ ) ( x 2 > x )
( ∀x ∈ ℜ )( ∃y ∈ℜ )( y = sin x )
( ∀x ∈ ℜ )( ∃y ∈ℜ )( y = tan x )
( ∃x ∈ ℜ )( ∃y ∈ℜ )( y = sin x )
( ∃x, y ∈ )( ∃n > 2 ) ( x n + y n = 1)
( ∃x ∈ ℜ )( ∀a, b, c ∈ℜ ) ( ax 2 + bx + c = 0 )
( ∃x ∈ )( ∀a, b, c ∈ℜ ) ( ax 2 + bx + c = 0 )
j)
( ∀ε > 0 )( ∃N ∈ )( ∀n > N ) 
k)
( ∀ε > 0 )( ∃δ > 0 ) ( x − 2 < δ ⇒
b)
c)
d)
e)
f)
g)
h)
2.
1

<ε 
n

x2 − 4 < ε
)
(Predicate Logic Form) Write the following theorems as predicate logic
notation.
a)
b)
c)
A number is divisible by 4 if and only if its last two digits are.
A number is divisible by 2n if and only if its last n digits are.
There exists irrational numbers x, y such that x y is rataional.
d)
e = ∑ n =0 1
e)
∑
f)
For positive real numbers a, b , we have
g)
If a, b are integers and b ≠ 0 then there exists a unique integers q, r such
∞
n
k =1
k=
is irrational.
n!
n ( n + 1)
2
ab ≤
a+b
.
2
that a = qb + r where 0 ≤ r < b .
h) If p is a prime number that does not divide the integer a then p
divides a p − a .
i)
The square of any natural number must have each prime number occurring an
even number of times in the prime factorization of the number.
Section 1.5
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Proofs in Predicate Logic
All prime numbers x greater than 2 are either of the form x = 2k + 1 or
x = 2k + 3 for some natural number k .
j)
k) Every even number greater than 4 can be expressed as the sum or two prime
numbers.
3.
(Negation)
Negation) Negate the following theorems. Which is true, the original
statement or its negation? Let the universe of all variables be the real
numbers.
a) ( ∃x )( ∀y )( xy < 1)
b)
c)
d)
e)
f)
4.
( ∀x )( ∀y )( ∃z )( xyz = 1)
( ∀x )( ∀y )( ∀z )( ∃w ) ( x 2 + y 2 + z 2 + w2 = 0 )
∼ ( ∃x )( ∀y )( x < y )
( ∀x )( ∃y )( xy < 1 ∧ xy > 1)
( ∃x )( ∃y )( xy = 0 ∨ xy ≠ 0 )
(Counterexamples
Counterexamples)
ples)
For each of the following find a counterexample3.
a)
There are no positive integers x greater than 2 that satisfy4
xx = x .
b)
For all positive integers, x 2 + x + 41 is prime.
c)
Every continuous function defined on the interval (0,1) has a
maximum and minimum value.
d)
Every continuous function is differentiable.
e)
If the terms of an infinite series approach zero, then the
series converges.
f)
e x > x for all real x .
g)
( ∀x ∈ ℜ )( ∃y ∈ℜ ) ( x = y 2 )
3
A nice book outlining many counterexamples in mathematics is Counterexamples in Mathematics by
Bernard R. Gelbaum and John Olmsted (Holden Day, Inc), 1964.
4
A famous mathematician once said that he was once
famous mathematician?
x 2 years old in the year x . Can you name the
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Section 1.5
h)
i)
j)
Proofs in Predicate Logic
There is no largest natural number.
If f is continuous then f is differentiable.
If
{ f n : n = 1, 2,...}
is a sequence of continuous functions defined on
(0,1) that converge to a function f , then f is continuous.
5.
(If and Only If Theorems)
Theorems) State each of the following “if and only if”
theorems in symbolic predicate logic notation, and then state the negation in
symbolic notation. In each case we assume f is a real valued function of a
real variable.
a) A function f is even iff for every real number x , f ( x) = f (− x) .
b) A function f is off iff for every real number x , f ( x) = − f (− x) .
c) A function f is periodic iff there exists a p such that
f ( x) = f ( x + p ) for all real numbers x .
d) A function is increasing iff for every real numbers x and y we have
x ≤ y ⇒ f ( x) ≤ f ( y ).
e) A function f is continuous at x0 iff for any ε > 0 there exists a
δ > 0 such that x − x0 < δ implies f ( x) − f ( x0 ) < ε .
f) A function f is uniformly continuous on a set E iff for any ε > 0 there
exists a δ > 0 such that
x − y < δ ..
f ( x) − f ( y ) < ε for any x, y in E that satisfy