This document provides an overview of mathematical proofs, emphasizing their correctness and completeness as essential for establishing truths in mathematics. It discusses various proof methods, terminology such as theorems, lemmas, and fallacies, and includes examples of formal proofs and inference rules. Additionally, it highlights the significance of proofs in logical argumentation and its applications in fields like program verification and computer security.

1. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 1
Module #2:
Logic: Basic Proof Methods
Rosen 5
Rosen 5th
th
ed., §1.5
ed., §1.5
48 slides, ~3 lectures
48 slides, ~3 lectures

2. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 2
Nature & Importance of Proofs
• In mathematics, a
In mathematics, a proof
proof is:
is:
– a
a correct
correct (well-reasoned, logically valid) and
(well-reasoned, logically valid) and complete
complete
(clear, detailed) argument that rigorously & undeniably
(clear, detailed) argument that rigorously & undeniably
establishes the truth of a mathematical statement.
establishes the truth of a mathematical statement.
• Why must the argument be correct & complete?
Why must the argument be correct & complete?
– Correctness
Correctness prevents us from fooling ourselves.
prevents us from fooling ourselves.
– Completeness
Completeness allows anyone to verify the result.
allows anyone to verify the result.
• In this course (& throughout mathematics), a
In this course (& throughout mathematics), a very
very
high standard
high standard for correctness and completeness of
for correctness and completeness of
proofs is demanded!!
proofs is demanded!!

3. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 3
Overview of §1.5
• Methods of mathematical argument (
Methods of mathematical argument (i.e.
i.e.,
,
proof methods) can be formalized in terms
proof methods) can be formalized in terms
of
of rules of logical inference
rules of logical inference.
.
• Mathematical
Mathematical proofs
proofs can themselves be
can themselves be
represented formally as discrete structures.
represented formally as discrete structures.
• We will review both
We will review both correct
correct &
& fallacious
fallacious
inference rules, & several proof methods.
inference rules, & several proof methods.

4. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 4
Applications of Proofs
• An exercise in clear communication of logical
An exercise in clear communication of logical
arguments in any area of study.
arguments in any area of study.
• The fundamental activity of mathematics is the
The fundamental activity of mathematics is the
discovery and elucidation, through proofs, of
discovery and elucidation, through proofs, of
interesting new theorems.
interesting new theorems.
• Theorem-proving has applications in program
Theorem-proving has applications in program
verification, computer security, automated
verification, computer security, automated
reasoning systems,
reasoning systems, etc.
etc.
• Proving a theorem allows us to rely upon on its
Proving a theorem allows us to rely upon on its
correctness even in the most critical scenarios.
correctness even in the most critical scenarios.

5. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 5
Proof Terminology
• Theorem
Theorem
– A statement that has been proven to be true.
A statement that has been proven to be true.
• Axioms
Axioms,
, postulates
postulates,
, hypotheses
hypotheses,
, premises
premises
– Assumptions (often unproven) defining the
Assumptions (often unproven) defining the
structures about which we are reasoning.
structures about which we are reasoning.
• Rules of inference
Rules of inference
– Patterns of logically valid deductions from
Patterns of logically valid deductions from
hypotheses to conclusions.
hypotheses to conclusions.

6. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 6
More Proof Terminology
• Lemma
Lemma -
- A minor theorem used as a stepping-
A minor theorem used as a stepping-
stone to proving a major theorem.
stone to proving a major theorem.
• Corollary
Corollary -
- A minor theorem proved as an easy
A minor theorem proved as an easy
consequence of a major theorem.
consequence of a major theorem.
• Conjecture
Conjecture -
- A statement whose truth value has
A statement whose truth value has
not been proven.
not been proven. (A conjecture may be widely
(A conjecture may be widely
believed to be true, regardless.)
believed to be true, regardless.)
• Theory
Theory –
– The set of all theorems that can be
The set of all theorems that can be
proven from a given set of axioms.
proven from a given set of axioms.

7. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 7
Graphical Visualization
…
Various Theorems
Various Theorems
The Axioms
The Axioms
of the Theory
of the Theory
A Particular Theory
A Particular Theory
A proof
A proof

8. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 8
Inference Rules - General Form
• An
An Inference Rule
Inference Rule is
is
– A pattern establishing that if we know that a set
A pattern establishing that if we know that a set
of
of antecedent
antecedent statements of certain forms are all
statements of certain forms are all
true, then we can validly deduce that a certain
true, then we can validly deduce that a certain
related
related consequent
consequent statement is true.
statement is true.
• antecedent 1
antecedent 1
antecedent 2 …
antecedent 2 …

 consequent
consequent “
“
” means “therefore”
” means “therefore”

9. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 9
Inference Rules & Implications
• Each valid logical inference rule
Each valid logical inference rule
corresponds to an implication that is a
corresponds to an implication that is a
tautology.
tautology.
• antecedent 1
antecedent 1 Inference rule
Inference rule
antecedent 2 …
antecedent 2 …

 consequent
consequent
• Corresponding tautology:
Corresponding tautology:
((
((ante. 1
ante. 1)
) 
 (
(ante. 2
ante. 2)
) 
 …)
…) 
 consequent
consequent

10. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 10
Some Inference Rules
• p
p Rule of Addition
Rule of Addition

 p
p
q
q
• p
p
q
q Rule of Simplification
Rule of Simplification

 p
p
• p
p Rule of Conjunction
Rule of Conjunction
q
q

 p
p
q
q

11. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 11
Modus Ponens & Tollens
• p
p Rule of
Rule of modus ponens
modus ponens
p
p
q
q (a.k.a.
(a.k.a. law of detachment
law of detachment)
)

q
q
• 
q
q
p
p
q
q Rule of
Rule of modus tollens
modus tollens

p
p
“the mode of
affirming”
“the mode of denying”

12. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 12
Syllogism Inference Rules
• p
p
q
q Rule of hypothetical
Rule of hypothetical
q
q
r
r syllogism
syllogism

p
p
r
r
• p
p 
 q
q Rule of disjunctive
Rule of disjunctive

p
p syllogism
syllogism

 q
q
Aristotle
(ca. 384-322 B.C.)

13. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 13
Formal Proofs
• A formal proof of a conclusion
A formal proof of a conclusion C
C, given
, given
premises
premises p
p1
1,
, p
p2
2,…
,…,
,p
pn
n consists of a sequence
consists of a sequence
of
of steps
steps, each of which applies some
, each of which applies some
inference rule to premises or previously-
inference rule to premises or previously-
proven statements (
proven statements (antecedents
antecedents) to yield a
) to yield a
new true statement (the
new true statement (the consequent
consequent).
).
• A proof demonstrates that
A proof demonstrates that if
if the premises are
the premises are
true,
true, then
then the conclusion is true.
the conclusion is true.

14. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 14
Formal Proof Example
• Suppose we have the following premises:
Suppose we have the following premises:
“It is not sunny and it is cold.”
“It is not sunny and it is cold.”
“We will swim only if it is sunny.”
“We will swim only if it is sunny.”
“If we do not swim, then we will canoe.”
“If we do not swim, then we will canoe.”
“If we canoe, then we will be home early.”
“If we canoe, then we will be home early.”
• Given these premises, prove the theorem
Given these premises, prove the theorem
“We will be home early”
“We will be home early” using inference rules.
using inference rules.

15. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 15
Proof Example cont.
• Let us adopt the following abbreviations:
Let us adopt the following abbreviations:
– sunny
sunny =
= “It is sunny”
“It is sunny”;
; cold
cold =
= “It is cold”
“It is cold”;
;
swim
swim =
= “We will swim”
“We will swim”;
; canoe
canoe =
= “We will
“We will
canoe”
canoe”;
; early
early =
= “We will be home early”
“We will be home early”.
.
• Then, the premises can be written as:
Then, the premises can be written as:
(1)
(1) 
sunny
sunny 
 cold
cold (2)
(2) swim
swim 
 sunny
sunny
(3)
(3) 
swim
swim 
 canoe
canoe (4)
(4) canoe
canoe 
 early
early

16. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 16
Proof Example cont.
Step
Step Proved by
Proved by
1.
1. 
sunny
sunny 
 cold
cold Premise #1.
Premise #1.
2.
2. 
sunny
sunny Simplification of 1.
Simplification of 1.
3.
3. swim
swim
sunny
sunny Premise #2.
Premise #2.
4.
4. 
swim
swim Modus tollens on 2,3.
Modus tollens on 2,3.
5.
5. 
swim
swim
canoe
canoe Premise #3.
Premise #3.
6.
6. canoe
canoe Modus ponens on 4,5.
Modus ponens on 4,5.
7.
7. canoe
canoe
early
early Premise #4.
Premise #4.
8.
8. early
early Modus ponens on 6,7.
Modus ponens on 6,7.

17. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 17
Inference Rules for Quantifiers
 
x
x P
P(
(x
x)
)

P
P(
(o
o)
) (substitute
(substitute any
any specific object
specific object o
o)
)
• P
P(
(g
g)
) (for
(for g
g a
a general
general element of u.d.)
element of u.d.)

x
x P
P(
(x
x)
)
 
x
x P
P(
(x
x)
)

P
P(
(c
c)
) (substitute a
(substitute a new
new constant
constant c
c)
)
• P
P(
(o
o)
) (substitute any extant object
(substitute any extant object o
o)
)

x
x P
P(
(x
x)
)

18. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 18
Common Fallacies
• A
A fallacy
fallacy is an inference rule or other proof
is an inference rule or other proof
method that is not logically valid.
method that is not logically valid.
– A fallacy may yield a false conclusion!
A fallacy may yield a false conclusion!
• Fallacy of
Fallacy of affirming the conclusion
affirming the conclusion:
:
– “
“p
p
q
q is true, and
is true, and q
q is true, so
is true, so p
p must be true.”
must be true.”
(No, because
(No, because F
F
T
T is true.)
is true.)
• Fallacy of
Fallacy of denying the hypothesis
denying the hypothesis:
:
– “
“p
p
q
q is true, and
is true, and p
p is false, so
is false, so q
q must be false.”
must be false.”
(No, again because
(No, again because F
F
T
T is true.)
is true.)

19. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 19
Circular Reasoning
• The fallacy of (explicitly or implicitly) assuming
The fallacy of (explicitly or implicitly) assuming
the very statement you are trying to prove in the
the very statement you are trying to prove in the
course of its proof. Example:
course of its proof. Example:
• Prove that an integer
Prove that an integer n
n is even, if
is even, if n
n2
2
is even.
is even.
• Attempted proof:
Attempted proof: “Assume
“Assume n
n2
2
is even. Then
is even. Then
n
n2
2
=2
=2k
k for some integer
for some integer k
k. Dividing both sides by
. Dividing both sides by n
n
gives
gives n
n = (2
= (2k
k)/
)/n
n = 2(
= 2(k
k/
/n
n)
). So there is an integer
. So there is an integer j
j
(namely
(namely k
k/
/n
n) such that
) such that n
n=2
=2j
j. Therefore
. Therefore n
n is even.”
is even.”
– Circular reasoning is used in this proof. Where?
Circular reasoning is used in this proof. Where?
Begs the question: How do
you show that j=k/n=n/2 is an integer,
without first assuming that n is even?

20. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 20
A Correct Proof
We know that
We know that n
n must be either odd or even.
must be either odd or even.
If
If n
n were odd, then
were odd, then n
n2
2
would be odd, since an
would be odd, since an
odd number times an odd number is always
odd number times an odd number is always
an odd number. Since
an odd number. Since n
n2
2
is even, it is not odd,
is even, it is not odd,
since no even number is also an odd number.
since no even number is also an odd number.
Thus, by modus tollens,
Thus, by modus tollens, n
n is not odd either.
is not odd either.
Thus, by disjunctive syllogism,
Thus, by disjunctive syllogism, n
n must be
must be
even.
even. ■
■ This proof is correct, but not quite complete,
since we used several lemmas without proving
them. Can you identify what they are?

21. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 21
A More Verbose Version
Suppose
Suppose n
n2
2
is even
is even 
2|
2|n
n2
2

 n
n2
2
mod 2 = 0. Of course
mod 2 = 0. Of course
n
n mod 2 is either 0 or 1. If it’s 1, then
mod 2 is either 0 or 1. If it’s 1, then n
n
1 (mod 2),
1 (mod 2),
so
so n
n2
2

1 (mod 2),
1 (mod 2), using the theorem that if
using the theorem that if a
a
b
b
(mod
(mod m
m) and
) and c
c
d
d (mod
(mod m
m) then
) then ac
ac
bd
bd (mod m),
(mod m),
with
with a
a=
=c
c=
=n
n and
and b
b=
=d
d=1.
=1. Now
Now n
n2
2

1 (mod 2) implies
1 (mod 2) implies
that
that n
n2
2
mod 2 = 1. So
mod 2 = 1. So by the hypothetical syllogism
by the hypothetical syllogism
rule
rule, (
, (n
n mod 2 = 1) implies (
mod 2 = 1) implies (n
n2
2
mod 2 = 1). Since we
mod 2 = 1). Since we
know
know n
n2
2
mod 2 = 0
mod 2 = 0 
 1,
1, by
by modus tollens
modus tollens we know
we know
that
that n
n mod 2
mod 2 
 1. So
1. So by disjunctive syllogism
by disjunctive syllogism we
we
have that
have that n
n mod 2 = 0
mod 2 = 0 
2|
2|n
n 
 n
n is even.
is even.
Uses some number theory we haven’t defined yet.

22. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 22
Proof Methods for Implications
For proving implications
For proving implications p
p
q
q, we have:
, we have:
• Direct
Direct proof:
proof: Assume
Assume p
p is true, and prove
is true, and prove q
q.
.
• Indirect
Indirect proof:
proof: Assume
Assume 
q
q, and prove
, and prove 
p
p.
.
• Vacuous
Vacuous proof:
proof: Prove
Prove 
p
p by itself.
by itself.
• Trivial
Trivial proof:
proof: Prove
Prove q
q by itself.
by itself.
• Proof by cases:
Proof by cases:
Show
Show p
p
(
(a
a 
 b
b), and (
), and (a
a
q
q) and (
) and (b
b
q
q).
).

23. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 23
Direct Proof Example
• Definition:
Definition: An integer
An integer n
n is called
is called odd
odd iff
iff n
n=2
=2k
k+1
+1
for some integer
for some integer k
k;
; n
n is
is even
even iff
iff n
n=2
=2k
k for some
for some k
k.
.
• Theorem:
Theorem: Every integer is either odd or even.
Every integer is either odd or even.
– This can be proven from even simpler axioms.
This can be proven from even simpler axioms.
• Theorem:
Theorem: (For all numbers
(For all numbers n
n) If
) If n
n is an odd
is an odd
integer, then
integer, then n
n2
2
is an odd integer.
is an odd integer.
• Proof:
Proof: If
If n
n is odd, then
is odd, then n
n = 2
= 2k
k+1
+1 for some integer
for some integer
k
k. Thus,
. Thus, n
n2
2
= (2
= (2k
k+1)
+1)2
2
= 4
= 4k
k2
2
+ 4
+ 4k
k + 1 = 2(2
+ 1 = 2(2k
k2
2
+ 2
+ 2k
k)
)
+ 1
+ 1. Therefore
. Therefore n
n2
2
is of the form
is of the form 2
2j
j + 1
+ 1 (with
(with j
j the
the
integer
integer 2
2k
k2
2
+ 2
+ 2k
k), thus
), thus n
n2
2
is odd.
is odd. □
□

24. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 24
Indirect Proof Example
• Theorem:
Theorem: (For all integers
(For all integers n
n)
)
If
If 3
3n
n+2
+2 is odd, then
is odd, then n
n is odd.
is odd.
• Proof:
Proof: Suppose that the conclusion is false,
Suppose that the conclusion is false, i.e.
i.e.,
,
that
that n
n is even. Then
is even. Then n
n=2
=2k
k for some integer
for some integer k
k.
.
Then
Then 3
3n
n+2 = 3(2
+2 = 3(2k
k)+2 = 6
)+2 = 6k
k+2 = 2(3
+2 = 2(3k
k+1)
+1). Thus
. Thus
3
3n
n+2
+2 is even, because it equals
is even, because it equals 2
2j
j for integer
for integer j
j =
=
3
3k
k+1
+1. So
. So 3
3n
n+2
+2 is not odd. We have shown that
is not odd. We have shown that
¬(
¬(n
n is odd)→¬(3
is odd)→¬(3n
n+2 is odd)
+2 is odd), thus its contra-
, thus its contra-
positive
positive (3
(3n
n+2 is odd) → (
+2 is odd) → (n
n is odd)
is odd) is also true. □
is also true. □

25. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 25
Vacuous Proof Example
• Theorem:
Theorem: (For all
(For all n
n) If
) If n
n is both odd and
is both odd and
even, then
even, then n
n2
2
=
= n
n +
+ n
n.
.
• Proof:
Proof: The statement “
The statement “n
n is both odd and
is both odd and
even” is necessarily false, since no number
even” is necessarily false, since no number
can be both odd and even. So, the theorem
can be both odd and even. So, the theorem
is vacuously true.
is vacuously true. □
□

26. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 26
Trivial Proof Example
• Theorem:
Theorem: (For integers
(For integers n
n) If
) If n
n is the sum
is the sum
of two prime numbers
of two prime numbers, then either
, then either n
n is odd
is odd
or
or n
n is even.
is even.
• Proof:
Proof: Any
Any integer
integer n
n is either odd or even.
is either odd or even.
So the conclusion of the implication is true
So the conclusion of the implication is true
regardless of the truth of the antecedent.
regardless of the truth of the antecedent.
Thus the implication is true trivially. □
Thus the implication is true trivially. □

27. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 27
Proof by Contradiction
• A method for proving
A method for proving p
p.
.
• Assume
Assume 
p
p, and prove both
, and prove both q
q and
and 
q
q for some
for some
proposition
proposition q
q. (Can be anything!)
. (Can be anything!)
• Thus
Thus 
p
p
 (
(q
q 
 
q
q)
)
• (
(q
q 
 
q
q) is a trivial contradiction, equal to
) is a trivial contradiction, equal to F
F
• Thus
Thus 
p
p
F
F, which is only true if
, which is only true if 
p
p=
=F
F
• Thus
Thus p
p is true.
is true.

28. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 28
Proof by Contradiction Example
• Theorem:
Theorem: is irrational.
is irrational.
– Proof:
Proof: Assume 2
Assume 21/2
1/2
were rational. This means
were rational. This means
there are integers
there are integers i
i,
,j
j with no common divisors
with no common divisors
such that 2
such that 21/2
1/2
=
= i
i/
/j
j. Squaring both sides, 2 =
. Squaring both sides, 2 =
i
i2
2
/
/j
j2
2
, so 2
, so 2j
j2
2
=
= i
i2
2
. So
. So i
i2
2
is even; thus
is even; thus i
i is even.
is even.
Let
Let i
i=2
=2k
k. So 2
. So 2j
j2
2
= (2
= (2k
k)
)2
2
= 4
= 4k
k2
2
. Dividing both
. Dividing both
sides by 2,
sides by 2, j
j2
2
= 2
= 2k
k2
2
. Thus
. Thus j
j2
2
is even, so
is even, so j
j is even.
is even.
But then
But then i
i and
and j
j have a common divisor,
have a common divisor,
namely 2, so we have a contradiction.
namely 2, so we have a contradiction. □
□
2 2

29. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 29
Review: Proof Methods So Far
• Direct
Direct,
, indirect
indirect,
, vacuous
vacuous, and
, and trivial
trivial proofs
proofs
of statements of the form
of statements of the form p
p
q
q.
.
• Proof by contradiction
Proof by contradiction of any statements.
of any statements.
• Next:
Next: Constructive
Constructive and
and nonconstructive
nonconstructive
existence proofs
existence proofs.
.

30. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 30
Proving Existentials
• A proof of a statement of the form
A proof of a statement of the form 
x
x P
P(
(x
x)
)
is called an
is called an existence proof
existence proof.
.
• If the proof demonstrates how to actually
If the proof demonstrates how to actually
find or construct a specific element
find or construct a specific element a
a such
such
that
that P
P(
(a
a) is true, then it is a
) is true, then it is a constructive
constructive
proof.
proof.
• Otherwise, it is
Otherwise, it is nonconstructive
nonconstructive.
.

31. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 31
Constructive Existence Proof
• Theorem:
Theorem: There exists a positive integer
There exists a positive integer n
n
that is the sum of two perfect cubes in two
that is the sum of two perfect cubes in two
different ways:
different ways:
– equal to
equal to j
j3
3
+
+ k
k3
3
and
and l
l3
3
+
+ m
m3
3
where
where j
j,
, k
k,
, l
l,
, m
m are
are
positive integers, and {
positive integers, and {j
j,
,k
k}
} ≠ {
≠ {l
l,
,m
m}
}
• Proof:
Proof: Consider
Consider n
n = 1729,
= 1729, j
j = 9,
= 9, k
k = 10,
= 10,
l
l = 1,
= 1, m
m = 12. Now just check that the
= 12. Now just check that the
equalities hold.
equalities hold.

32. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 32
Another Constructive
Existence Proof
• Theorem:
Theorem: For any integer
For any integer n
n>0, there exists
>0, there exists
a sequence of
a sequence of n
n consecutive composite
consecutive composite
integers.
integers.
• Same statement in predicate logic:
Same statement in predicate logic:

n
n>0
>0 
x
x 
i
i (1
(1
i
i
n
n)
)
(
(x
x+
+i
i is composite)
is composite)
• Proof follows on next slide…
Proof follows on next slide…

33. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 33
The proof...
• Given
Given n
n>0, let
>0, let x
x = (
= (n
n + 1)! + 1.
+ 1)! + 1.
• Let
Let i
i 
 1 and
1 and i
i 
 n
n, and consider
, and consider x
x+
+i
i.
.
• Note
Note x
x+
+i
i = (
= (n
n + 1)! + (
+ 1)! + (i
i + 1).
+ 1).
• Note (
Note (i
i+1)|(
+1)|(n
n+1)!, since 2
+1)!, since 2 
 i
i+1
+1 
 n
n+1.
+1.
• Also (
Also (i
i+1)|(
+1)|(i
i+1). So, (
+1). So, (i
i+1)|(
+1)|(x+i
x+i).
).
 
 x+i
x+i is composite.
is composite.
 
 
n
n 
x
x 
1
1
i
i
n
n :
: x
x+
+i
i is composite. Q.E.D.
is composite. Q.E.D.

34. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 34
Nonconstructive Existence Proof
• Theorem:
Theorem:
“There are infinitely many prime numbers.”
“There are infinitely many prime numbers.”
• Any finite set of numbers must contain a maximal
Any finite set of numbers must contain a maximal
element, so we can prove the theorem if we can
element, so we can prove the theorem if we can
just show that there is
just show that there is no
no largest prime number.
largest prime number.
• I.e.
I.e., show that for any prime number, there is a
, show that for any prime number, there is a
larger number that is
larger number that is also
also prime.
prime.
• More generally: For
More generally: For any
any number,
number, 
 a larger prime.
a larger prime.
• Formally: Show
Formally: Show 
n
n 
p>n
p>n :
: p
p is prime.
is prime.

35. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 35
The proof, using proof by cases...
• Given
Given n
n>0, prove there is a prime
>0, prove there is a prime p
p>
>n
n.
.
• Consider
Consider x
x =
= n
n!+1. Since
!+1. Since x
x>1, we know
>1, we know
(
(x
x is prime)
is prime)
(
(x
x is composite).
is composite).
• Case 1:
Case 1: x
x is prime. Obviously
is prime. Obviously x
x>
>n
n, so let
, so let
p
p=
=x
x and we’re done.
and we’re done.
• Case 2:
Case 2: x
x has a prime factor
has a prime factor p
p. But if
. But if p
p
n
n,
,
then
then p
p mod
mod x
x = 1. So
= 1. So p
p>
>n
n, and we’re done.
, and we’re done.

36. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 36
The Halting Problem (Turing‘36)
• The
The halting problem
halting problem was the first
was the first
mathematical function proven to
mathematical function proven to
have
have no
no algorithm that computes it!
algorithm that computes it!
– We say, it is
We say, it is uncomputable.
uncomputable.
• The desired function is
The desired function is Halts
Halts(
(P
P,
,I
I) :
) :≡
≡
the truth value of this statement:
the truth value of this statement:
– “
“Program P, given input I, eventually terminates.”
Program P, given input I, eventually terminates.”
• Theorem:
Theorem: Halts
Halts is uncomputable!
is uncomputable!
– I.e., There does
I.e., There does not
not exist
exist any
any algorithm
algorithm A
A that
that
computes
computes Halts
Halts correctly for
correctly for all
all possible inputs.
possible inputs.
• Its proof is thus a
Its proof is thus a non
non-existence proof.
-existence proof.
• Corollary:
Corollary: General impossibility of predictive analysis of
General impossibility of predictive analysis of
arbitrary computer programs.
arbitrary computer programs.
Alan Turing
1912-1954

37. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 37
The Proof
• Given any
Given any arbitrary
arbitrary program
program H
H(
(P,I
P,I),
),
• Consider algorithm
Consider algorithm Breaker
Breaker, defined as:
, defined as:
procedure
procedure Breaker
Breaker(
(P
P: a program)
: a program)
halts
halts :=
:= H
H(
(P
P,
,P
P)
)
if
if halts
halts then while
then while T begin end
T begin end
• Note that
Note that Breaker
Breaker(
(Breaker
Breaker) halts iff
) halts iff
H
H(
(Breaker
Breaker,
,Breaker
Breaker) =
) = F
F.
.
• So
So H
H does
does not
not compute the function
compute the function Halts
Halts!
!
Breaker makes a
liar out of H, by
doing the opposite
of whatever H
predicts.

38. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 38
Limits on Proofs
• Some very simple statements of number
Some very simple statements of number
theory haven’t been proved or disproved!
theory haven’t been proved or disproved!
– E.g. Goldbach’s conjecture
E.g. Goldbach’s conjecture: Every integer
: Every integer n
n≥
≥2
2
is exactly the average of some two primes.
is exactly the average of some two primes.
 
n
n≥
≥2
2 
 primes
primes p
p,
,q
q:
: n
n=(
=(p
p+
+q
q)/2.
)/2.
• There are true statements of number theory
There are true statements of number theory
(or any sufficiently powerful system) that
(or any sufficiently powerful system) that
can
can never
never be proved (or disproved) (Gödel).
be proved (or disproved) (Gödel).

39. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 39
More Proof Examples
• Quiz question 1a: Is this argument correct or
Quiz question 1a: Is this argument correct or
incorrect?
incorrect?
– “
“All TAs compose easy quizzes. Ramesh is a TA.
All TAs compose easy quizzes. Ramesh is a TA.
Therefore, Ramesh composes easy quizzes.”
Therefore, Ramesh composes easy quizzes.”
• First, separate the premises from conclusions:
First, separate the premises from conclusions:
– Premise #1: All TAs compose easy quizzes.
Premise #1: All TAs compose easy quizzes.
– Premise #2: Ramesh is a TA.
Premise #2: Ramesh is a TA.
– Conclusion: Ramesh composes easy quizzes.
Conclusion: Ramesh composes easy quizzes.

40. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 40
Answer
Next, re-render the example in logic notation.
Next, re-render the example in logic notation.
• Premise #1: All TAs compose easy quizzes.
Premise #1: All TAs compose easy quizzes.
– Let U.D. = all people
Let U.D. = all people
– Let
Let T
T(
(x
x) :
) :≡ “
≡ “x
x is a TA”
is a TA”
– Let
Let E
E(
(x
x) :≡ “
) :≡ “x
x composes easy quizzes”
composes easy quizzes”
– Then Premise #1 says:
Then Premise #1 says: 
x
x,
, T
T(
(x
x)→
)→E
E(
(x
x)
)

41. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 41
Answer cont…
• Premise #2: Ramesh is a TA.
Premise #2: Ramesh is a TA.
– Let R :
Let R :≡ Ramesh
≡ Ramesh
– Then Premise #2 says:
Then Premise #2 says: T
T(R)
(R)
– And the Conclusion says:
And the Conclusion says: E
E(R)
(R)
• The argument is correct, because it can be
The argument is correct, because it can be
reduced to a sequence of applications of
reduced to a sequence of applications of
valid inference rules, as follows:
valid inference rules, as follows:

42. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 42
The Proof in Gory Detail
• Statement
Statement How obtained
How obtained
1.
1. 
x
x,
, T
T(
(x
x)
) →
→ E
E(
(x
x)
) (Premise #1)
(Premise #1)
2.
2. T
T(Ramesh) →
(Ramesh) → E
E(Ramesh)
(Ramesh) (Universal
(Universal
instantiation)
instantiation)
3.
3. T
T(Ramesh)
(Ramesh) (Premise #2)
(Premise #2)
4.
4. E
E(Ramesh)
(Ramesh)(
(Modus Ponens
Modus Ponens from
from
statements #2 and #3)
statements #2 and #3)

43. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 43
Another example
• Quiz question 2b: Correct or incorrect: At least
Quiz question 2b: Correct or incorrect: At least
one of the 280 students in the class is intelligent.
one of the 280 students in the class is intelligent.
Y is a student of this class. Therefore, Y is
Y is a student of this class. Therefore, Y is
intelligent.
intelligent.
• First: Separate premises/conclusion,
First: Separate premises/conclusion,
& translate to logic:
& translate to logic:
– Premises: (1)
Premises: (1) 
x
x InClass(
InClass(x
x)
) 
 Intelligent(
Intelligent(x
x)
)
(2) InClass(Y)
(2) InClass(Y)
– Conclusion: Intelligent(Y)
Conclusion: Intelligent(Y)

44. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 44
Answer
• No, the argument is invalid; we can disprove it
No, the argument is invalid; we can disprove it
with a counter-example, as follows:
with a counter-example, as follows:
• Consider a case where there is only one intelligent
Consider a case where there is only one intelligent
student X in the class, and X
student X in the class, and X≠Y.
≠Y.
– Then the premise
Then the premise 
x
x InClass(
InClass(x
x)
) 
 Intelligent(
Intelligent(x
x)
) is true,
is true,
by existential generalization of
by existential generalization of
InClass(X)
InClass(X) 
 Intelligent(X)
Intelligent(X)
– But the conclusion
But the conclusion Intelligent(Y)
Intelligent(Y) is false, since X is
is false, since X is
the only intelligent student in the class, and Y
the only intelligent student in the class, and Y≠X.
≠X.
• Therefore, the premises
Therefore, the premises do not
do not imply the
imply the
conclusion.
conclusion.

45. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 45
Another Example
• Quiz question #2: Prove that the sum of a rational
Quiz question #2: Prove that the sum of a rational
number and an irrational number is always
number and an irrational number is always
irrational.
irrational.
• First, you have to understand exactly what the
First, you have to understand exactly what the
question is asking you to prove:
question is asking you to prove:
– “
“For all
For all real numbers
real numbers x
x,
,y,
y, if
if x
x is rational and
is rational and y
y is
is
irrational, then
irrational, then x
x+
+y
y is irrational.”
is irrational.”
 
x
x,
,y
y: Rational(
: Rational(x
x)
) 
 Irrational(
Irrational(y
y) → Irrational(
) → Irrational(x
x+
+y
y)
)

46. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 46
Answer
• Next, think back to the definitions of the
Next, think back to the definitions of the
terms used in the statement of the theorem:
terms used in the statement of the theorem:
 
 reals
reals r
r: Rational(
: Rational(r
r)
) ↔
↔

 Integer(
Integer(i
i)
) 
 Integer(
Integer(j
j):
): r
r =
= i
i/
/j
j.
.
 
 reals
reals r
r: Irrational(
: Irrational(r
r)
) ↔ ¬
↔ ¬Rational(
Rational(r
r)
)
• You almost always need the definitions of
You almost always need the definitions of
the terms in order to prove the theorem!
the terms in order to prove the theorem!
• Next, let’s go through one valid proof:
Next, let’s go through one valid proof:

47. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 47
What you might write
• Theorem:
Theorem:

x
x,
,y
y: Rational(
: Rational(x
x)
) 
 Irr
Irrational(
ational(y
y) Irrational(
→
) Irrational(
→ x
x+
+y
y)
)
• Proof:
Proof: Let
Let x
x,
, y
y be any rational and irrational
be any rational and irrational
numbers, respectively
numbers, respectively. … (universal generalization)
. … (universal generalization)
• Now, just from this, what do we know about
Now, just from this, what do we know about x
x and
and y
y? You
? You
should think back to the definition of rational:
should think back to the definition of rational:
• …
… Since
Since x
x is rational, we know (from the very
is rational, we know (from the very
definition of rational) that there must be some
definition of rational) that there must be some
integers
integers i
i and
and j
j such that
such that x
x =
= i
i/
/j
j.
. So, let
So, let i
ix
x,
,j
jx
x be
be
such integers
such integers …
…
• We give them unique names so we can refer to them later.
We give them unique names so we can refer to them later.

48. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 48
What next?
• What do we know about
What do we know about y
y? Only that
? Only that y
y is
is
irrational:
irrational: ¬
¬
 integers
integers i
i,
,j
j:
: y
y =
= i
i/
/j
j.
.
• But, it’s difficult to see how to use a direct proof
But, it’s difficult to see how to use a direct proof
in this case. We could try indirect proof also, but
in this case. We could try indirect proof also, but
in this case, it is a little simpler to just use proof
in this case, it is a little simpler to just use proof
by contradiction (very similar to indirect).
by contradiction (very similar to indirect).
• So, what are we trying to show? Just that
So, what are we trying to show? Just that x
x+
+y
y is
is
irrational. That is,
irrational. That is, ¬
¬
i
i,
,j
j: (
: (x
x +
+ y
y) =
) = i
i/
/j
j.
.
• What happens if we hypothesize the negation of
What happens if we hypothesize the negation of
this statement?
this statement?

49. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 49
More writing…
• Suppose that
Suppose that x
x+
+y
y were not irrational. Then
were not irrational. Then
x
x+
+y
y would be rational, so
would be rational, so 
 integers
integers i
i,
,j
j:
: x
x+
+y
y
=
= i
i/
/j
j. So, let
. So, let i
is
s and
and j
js
s be any such integers
be any such integers
where
where x
x+
+y
y =
= i
is
s/
/ j
js
s.
.
• Now, with all these things named, we can start
Now, with all these things named, we can start
seeing what happens when we put them together.
seeing what happens when we put them together.
• So, we have that (
So, we have that (i
ix
x/
/j
jx
x) +
) + y
y = (
= (i
is
s/
/j
js
s).
).
• Observe! We have enough information now that
Observe! We have enough information now that
we can conclude something useful about
we can conclude something useful about y
y, by
, by
solving this equation for it.
solving this equation for it.

50. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 50
Finishing the proof.
• Solving that equation for
Solving that equation for y
y, we have:
, we have:
y
y = (
= (i
is
s/
/j
js
s) – (
) – (i
ix
x/
/j
jx
x)
)
= (
= (i
is
sj
jx
x –
– i
ix
xj
js
s)/(
)/(j
js
sj
jx
x)
)
Now, since the numerator and denominator
Now, since the numerator and denominator
of this expression are both integers,
of this expression are both integers, y
y is
is
(by definition) rational. This contradicts
(by definition) rational. This contradicts
the assumption that
the assumption that y
y was irrational.
was irrational.
Therefore, our hypothesis that
Therefore, our hypothesis that x
x+
+y
y is
is
rational must be false, and so the theorem
rational must be false, and so the theorem
is proved.
is proved.

51. Module #2 - Proofs
08/31/24 (c)2001-2003, Michael P. Frank 51
Example wrong answer
• 1 is rational. √2 is irrational. 1+√2 is
1 is rational. √2 is irrational. 1+√2 is
irrational. Therefore, the sum of a rational
irrational. Therefore, the sum of a rational
number and an irrational number is
number and an irrational number is
irrational. (Direct proof.)
irrational. (Direct proof.)
• Why does this answer desereve
Why does this answer desereve no credit
no credit?
?
– The student attempted to use an example to prove a
The student attempted to use an example to prove a
universal statement.
universal statement. This is
This is always invalid
always invalid!
!
– Even as an example, it’s incomplete, because the
Even as an example, it’s incomplete, because the
student never even proved that 1+
student never even proved that 1+√2 is irrational!
√2 is irrational!