Proof Review
CS 202
Aaron Bloomfield
Spring 2007
Proof methods in this course
• Logical equivalences
– via truth tables
– via logical equivalences
• Set equivalences
–
–
–
–
via set identities
via membership tables
via mutual subset proof
via set builder notation and
logical equivalences
• Rules of inference
– for propositions
– for quantified statements
• Proofs without a category
– Pigeonhole principle
– Combinatorial proofs
• Ten proof methods in chapter 3:
–
–
–
–
–
–
–
–
Direct proofs
Indirect proofs
Vacuous proofs
Trivial proofs
Proof by contradiction
Proof by cases
Proofs of equivalence
Existence proofs
• Constructive
• Non-constructive
– Uniqueness proofs
– Counterexamples
• Induction
– Weak mathematical induction
– Strong mathematical induction
– Structural induction
Logic proofs
Using Truth Tables
( p  r )  (q  r )  ( p  q)  r
p q r p→r q →r (p→r)(q →r) pq (pq) →r
T T T
T
T
T
T
T
T T F
F
F
F
T
F
T F T
T
T
T
F
T
T F F
F
T
T
F
T
F T T
T
T
T
F
T
F T F
T
F
T
F
T
F F T
T
T
T
F
T
F F F
T
T
T
F
T
Using Logical Equivalences
( p  r )  (q  r )  ( p  q)  r
Original statement
(p Definition
r )  (qofimplication
r )  ( p pq)
 qr  p  q
(p DeMorgan’s
r )  (q  Law
r )  (
r p  q
p( pqq))  
Associativity
p  r  q ofrOr (pprq) (rq  r )  p  r  q  r
Re-arranging
p  q  r  r  p  q  r
p  q  r  p  q  r
rr r
Idempotent Law
Set proofs
Proof by using basic set identities
• Prove that A∩B=B-(B-A)
A  B  B-(B  A )
Definition of difference
 B  (B  A )
Definition of difference
 B  (B  A )
DeMorgan’s law
 B  (B  A)
Complementation law
 (B  B )  (B  A)
Distributive law
   (B  A)
Complement law
 (B  A)
Identity law
 A B
Commutative law
Proof by membership tables
• The following membership table shows that
A∩B=B-(B-A)
A
1
1
0
0
B
1
0
1
0
A∩B
1
0
0
0
B-A
0
0
1
0
B-(B-A)
1
0
0
0
• Because the two indicated columns have the
same values, the two expressions are identical
• This is similar to Boolean logic!
Proof by showing each set
is a subset of the other 1
• Assume that xB-(B-A)
– By definition of difference, we know that xB and xB-A
• Consider xB-A
– If xB-A, then (by definition of difference) xB and xA
– Since xB-A, then only one of the inverses has to be true
(DeMorgan’s law): xB or xA
• So we have that xB and (xB or xA)
– It cannot be the case where xB and xB
– Thus, xB and xA
– This is the definition of intersection
• Thus, if xB-(B-A) then xA∩B
Proof by showing each set
is a subset of the other 2
• Assume that xA∩B
– By definition of intersection, xA and xB
• Thus, we know that xB-A
– B-A includes all the elements in B that are also not in A not
include any of the elements of A (by definition of difference)
• Consider B-(B-A)
– We know that xB-A
– We also know that if xA∩B then xB (by definition of
intersection)
– Thus, if xB and xB-A, we can restate that (using the definition
of difference) as xB-(B-A)
• Thus, if xA∩B then xB-(B-A)
Proof by set builder notation
and logical equivalences 2
B  ( B  A)
 {x | x  B  x  ( B  A)}
Original statement
 {x | x  B  ( x  ( B  A))}
Negating “element of”
 {x | x  B  ( x  B  x  A)}
Definition of difference
 {x | x  B  ( x  B  x  A)}
 {x | x  B  x  B  x  B  x  A}
DeMorgan’s Law
 {x | x  B  ( x  B)  x  B  x  A}
 {x | F  x  B  x  A}
 {x | x  B  x  A}
 A B
Definition of difference
Distributive Law
Negating “element of”
Negation Law
Identity Law
Definition of intersection
Rules of Inference
Modus Ponens example
• Assume you are given the following two
statements:
– “you are in this class”
– “if you are in this class, you will get a grade”
p
pq
q
• Let p = “you are in this class”
• Let q = “you will get a grade”
• By Modus Ponens, you can conclude that you
will get a grade
Modus Tollens
• Assume that we know: ¬q and p → q
– Recall that p → q = ¬q → ¬p
• Thus, we know ¬q and ¬q → ¬p
• We can conclude ¬p
q
pq
 p
Generalization & Specialization
• Generalization:
If
you know that p is
true, then p  q will
ALWAYS be true
• Specialization: If p 
q is true, then p will
ALWAYS be true
p
pq
pq
p
More rules of inference
• Conjunction: if p and q are
true separately, then pq is
true
p
q
pq
pq
• Elimination: If pq is true,
and p is false, then q must be
true
• Transitivity: If p→q is true,
and q→r is true, then p→r
must be true
p
q
pq
qr
pr
Even more rules of inference
• Proof by division into cases:
if at least one of p or q is
true, then r must be true
pq
pr
qr
r
• Contradiction rule: If ¬p→c is
true, we can conclude p (via
the contra-positive)
• Resolution: If pq is true, and
¬pr is true, then qr must
be true
– Not in the textbook
p  c
p
pq
p  r
q  r
Example of proof
•
• We have the hypotheses:
p
q
r
s
t
– “It is not sunny this afternoon and it
is colder than yesterday”
– “We will go swimming only if it is
sunny”
– “If we do not go swimming, then we
will take a canoe trip”
– “If we take a canoe trip, then we will
be home by sunset”
• Does this imply that “we will be
home by sunset”?
¬p  q
r→p
¬r → s
s→t
t
Example of proof
1.
2.
3.
4.
5.
6.
7.
8.
¬p  q
¬p
r→p
¬r
¬r → s
s
s→t
t
pq
p
1st hypothesis
Simplification using step 1
2nd hypothesis
Modus tollens using steps 2 & 3
3rd hypothesis
Modus ponens using steps 4 & 5
4th hypothesis
Modus ponens using steps 6 & 7
p
q
pq
pq
q
 p
Example of proof
• Given the hypotheses:
– “If it does not rain or if it is not
foggy, then the sailing race will
be held and the lifesaving
demonstration will go on”
– “If the sailing race is held, then
the trophy will be awarded”
– “The trophy was not awarded”
• Can you conclude: “It rained”?
(¬r  ¬f) →
(s  l)
s→t
¬t
r
Example of proof
1.
2.
3.
4.
5.
6.
7.
8.
9.
¬t
s→t
¬s
(¬r¬f)→(sl)
¬(sl)→¬(¬r¬f)
(¬s¬l)→(rf)
¬s¬l
rf
r
3rd hypothesis
2nd hypothesis
Modus tollens using steps 2 & 3
1st hypothesis
Contrapositive of step 4
DeMorgan’s law and double negation law
Addition from step 3
Modus ponens using steps 6 & 7
Simplification using step 8
q
p
pq
q
p
pq
pq
p
pq
 p
Proofs without a category
Corollary 1: algebraic proof
Let n be a non-negative integer. Then
n
n



2

 
k 0  k 
n
Algebraic proof
2n  (1  1) n
 n  k nk
   1 1
k 0  k 
n
n
   
k 0  k 
n
 n  n j j
( x  y )     x y
j 0  j 
n
n
Corollary 1: combinatorial proof
Let n be a non-negative integer. Then
n
n



2

 
k 0  k 
n
Combinatorial proof

A set with n elements has 2n subsets
By definition of power set

Each subset has either 0 or 1 or 2 or … or n elements
There are  n0  subsets with 0 elements,  1n  subsets with 1
 
 
n


element,
subsets with n elements
 … and
n
Thus, the total number of subsets is

Thus,
n
n



2

k 
k 0  
n
n
 

k 0  k 
n
Pigeonhole Principle
Consider 5 distinct points (xi, yi) with integer values, where i = 1,
2, 3, 4, 5
Show that the midpoint of at least one pair of these five points
also has integer coordinates
Thus, we are looking for the midpoint of a segment from (a,b) to
(c,d)

The midpoint is ( (a+c)/2, (b+d)/2 )
Note that the midpoint will be integers if a and c have the same
parity: are either both even or both odd

Same for b and d
There are four parity possibilities

(even, even), (even, odd), (odd, even), (odd, odd)
Since we have 5 points, by the pigeonhole principle, there must
be two points that have the same parity possibility

Thus, the midpoint of those two points will have integer coordinates
Ten proof methods in chapter 3
Direct proofs
• Consider an implication: p→q
– If p is false, then the implication is always true
– Thus, show that if p is true, then q is true
• To perform a direct proof, assume that p is
true, and show that q must therefore be
true
Direct proof example
• Show that the square of an even number
is an even number
– Rephrased: if n is even, then n2 is even
• Assume n is even
– Thus, n = 2k, for some k (definition of even
numbers)
– n2 = (2k)2 = 4k2 = 2(2k2)
– As n2 is 2 times an integer, n2 is thus even
Indirect proofs
• Consider an implication: p→q
– It’s contrapositive is ¬q→¬p
• Is logically equivalent to the original implication!
– If the antecedent (¬q) is false, then the
contrapositive is always true
– Thus, show that if ¬q is true, then ¬p is true
• To perform an indirect proof, do a direct
proof on the contrapositive
Indirect proof example
• If n2 is an odd integer then n is an odd integer
• Prove the contrapositive: If n is an even integer,
then n2 is an even integer
• Proof: n=2k for some integer k (definition of even
numbers)
• n2 = (2k)2 = 4k2 = 2(2k2)
• Since n2 is 2 times an integer, it is even
Vacuous proofs
• Consider an implication: p→q
• If it can be shown that p is false, then the
implication is always true
– By definition of an implication
• Note that you are showing that the
antecedent is false
Vacuous proof example
• Consider the statement:
– All criminology majors in CS 202 are female
– Rephrased: If you are a criminology major
and you are in CS 202, then you are female
• Could also use quantifiers!
• Since there are no criminology majors in
this class, the antecedent is false, and the
implication is true
Trivial proofs
• Consider an implication: p→q
• If it can be shown that q is true, then the
implication is always true
– By definition of an implication
• Note that you are showing that the
conclusion is true
Trivial proof example
• Consider the statement:
– If you are tall and are in CS 202 then you are
a student
• Since all people in CS 202 are students,
the implication is true regardless
Proof by contradiction
• Given a statement p, assume it is false
– Assume ¬p
• Prove that ¬p cannot occur
– A contradiction exists
• Given a statement of the form p→q
– To assume it’s false, you only have to consider the
case where p is true and q is false
Proof by contradiction example
• Theorem (by Euclid): There are infinitely many
prime numbers.
• Proof. Assume there are a finite number of primes
• List them as follows: p1, p2 …, pn.
• Consider the number q = p1p2 … pn + 1
– This number is not divisible by any of the listed primes
• If we divided pi into q, there would result a remainder of 1
– We must conclude that q is a prime number, not among
the primes listed above
• This contradicts our assumption that all primes are in the list
p1, p2 …, pn.
Proof by cases
• Show a statement is true by showing all
possible cases are true
• Thus, you are showing a statement of the
form:  p  p  ...  p   q
1
2
n
is true by showing that:
 p1  p2  ...  pn   q   p1  q    p2  q   ...   pn  q 
Proof by cases example
• Prove that
a a

b b
– Note that b ≠ 0
• Cases:
– Case 1: a ≥ 0 and b > 0
• Then |a| = a, |b| = b, and
– Case 2: a ≥ 0 and b < 0
• Then |a| = a, |b| = -b, and
– Case 3: a < 0 and b > 0
• Then |a| = -a, |b| = b, and
– Case 4: a < 0 and b < 0
• Then |a| = -a, |b| = -b, and
a a a
 
b b b
a
a
a
a
 

b
b b b
a
a a a
 

b
b
b
b
a a a a
 

b b b b
Proofs of equivalences
• This is showing the definition of a biconditional
• Given a statement of the form “p if and
only if q”
– Show it is true by showing (p→q)(q→p) is
true
Proofs of equivalence example
• Show that m2=n2 if and only if m=n or m=-n
– Rephrased: (m2=n2) ↔ [(m=n)(m=-n)]
• Need to prove two parts:
– [(m=n)(m=-n)] → (m2=n2)
• Proof by cases!
• Case 1: (m=n) → (m2=n2)
– (m)2 = m2, and (n)2 = n2, so this case is proven
• Case 2: (m=-n) → (m2=n2)
– (m)2 = m2, and (-n)2 = n2, so this case is proven
– (m2=n2) → [(m=n)(m=-n)]
•
•
•
•
Subtract n2 from both sides to get m2-n2=0
Factor to get (m+n)(m-n) = 0
Since that equals zero, one of the factors must be zero
Thus, either m+n=0 (which means m=n) or m-n=0 (which
means m=-n)
Existence proofs
• Given a statement: x P(x)
• We only have to show that a P(c) exists for
some value of c
• Two types:
– Constructive: Find a specific value of c for
which P(c) exists
– Nonconstructive: Show that such a c exists,
but don’t actually find it
• Assume it does not exist, and show a contradiction
Constructive existence proof
example
• Show that a square exists that is the sum
of two other squares
– Proof: 32 + 42 = 52
• Show that a cube exists that is the sum of
three other cubes
– Proof: 33 + 43 + 53 = 63
Non-constructive existence proof
example
• Prove that either 2*10500+15 or 2*10500+16 is not
a perfect square
– A perfect square is a square of an integer
– Rephrased: Show that a non-perfect square exists in
the set {2*10500+15, 2*10500+16}
• Proof: The only two perfect squares that differ by
1 are 0 and 1
– Thus, any other numbers that differ by 1 cannot both
be perfect squares
– Thus, a non-perfect square must exist in any set that
contains two numbers that differ by 1
– Note that we didn’t specify which one it was!
Uniqueness proofs
• A theorem may state that only one such
value exists
• To prove this, you need to show:
– Existence: that such a value does indeed
exist
• Either via a constructive or non-constructive
existence proof
– Uniqueness: that there is only one such value
Uniqueness proof example
• If the real number equation 5x+3=a has a
solution then it is unique
• Existence
– We can manipulate 5x+3=a to yield x=(a-3)/5
– Is this constructive or non-constructive?
• Uniqueness
– If there are two such numbers, then they would fulfill
the following: a = 5x+3 = 5y+3
– We can manipulate this to yield that x = y
• Thus, the one solution is unique!
Counterexamples
• Given a universally quantified statement, find a single
example which it is not true
• Note that this is DISPROVING a UNIVERSAL statement
by a counterexample
• x ¬R(x), where R(x) means “x has red hair”
– Find one person (in the domain) who has red hair
• Every positive integer is the square of another integer
– The square root of 5 is 2.236, which is not an integer
A note on counterexamples
• You can DISPROVE something by showing a
single counter example
– You are finding an example to show that something is
not true
• You cannot PROVE something by example
• Example: prove or disprove that all numbers are
even
– Proof by contradiction: 1 is not even
– (Invalid) proof by example: 2 is even
Induction
Mathematical induction example
• Inductive step: show
k 1
2
i
 
i 1
(k  1)(( k  1)  1)( 2(k  1)  1)
6
(k  1)(( k  1)  1)( 2(k  1)  1)
i 

6
i 1
k 1
2
k
(k  1) 2   i 2 
i 1
(k  1) 2 
(k  1)( k  2)( 2k  3)
6
k (k  1)( 2k  1) (k  1)( k  2)( 2k  3)

6
6
6(k  1) 2  k (k  1)( 2k  1)  (k  1)(k  2)(2k  3)
2k 3  9k 2  13k  6  2k 3  9k 2  13k  6
k
2
i
 
i 1
k (k  1)( 2k  1)
6
Strong induction example
• Show that any number > 1 can be written
as the product of one or more primes
• Base case: P(2)
– 2 is the product of 2 (remember that 1 is not
prime!)
• Inductive hypothesis: assume P(2), P(3),
…, P(k) are all true
• Inductive step: Show that P(k+1) is true
Strong induction example 1
• Inductive step: Show that P(k+1) is true
• There are two cases:
– k+1 is prime
• It can then be written as the product of k+1
– k+1 is composite
• It can be written as the product of two composites,
a and b, where 2 ≤ a ≤ b < k+1
• By the inductive hypothesis, both P(a) and P(b) are
true
Chess and induction
Can the knight reach any
square in a finite number
of moves?
7
6
5
Show that the knight can
reach any square (i, j) for
which i+j=k where k > 1.
4
Base case: k = 2
3
Inductive
hypothesis:
assume the knight can
reach any square (i, j) for
which i+j=k where k > 1.
2
Inductive step: show the
knight can reach any
square (i, j) for which
i+j=k+1 where k > 1.
1
0
0
1
2
3
4
5
6
7
Chess and induction
• Inductive step: show the knight can reach any
square (i, j) for which i+j=k+1 where k > 1.
– Note that k+1 ≥ 3, and one of i or j is ≥ 2
– If i ≥ 2, the knight could have moved from (i-2, j+1)
• Since i+j = k+1, i-2 + j+1 = k, which is assumed true
– If j ≥ 2, the knight could have moved from (i+1, j-2)
• Since i+j = k+1, i+1 + j-2 = k, which is assumed true