CS104 : Discrete Structures

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CS104 : Discrete Structures

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Title: CS104 : Discrete Structures

1
CS104 Discrete Structures
Chapter III Proof Techniques
2
Rules of Inference Valid Arguments in
Propositional Logic

Argument An argument is a sequence of statements
that end with a conclusion.
Valid An argument is valid if and only if it is
impossible for all premises (preceding
statements) to be true and the conclusion to be
false. By valid, we mean that the conclusion of
the argument must follow from the truth of the
premises of the argument.
Consider the arguments
If you have a current password, then you can
log onto the network.
You have a current password.
Therefore,
You can log onto the network.

3
Rules of Inference Valid Arguments in
Propositional Logic

The conclusion You can log onto the network
must be true when the premises If you have a
current password, then you can log onto the
network and You have a current password are
true.
Let p You have a current password.
and qYou can log onto the network. Then the
argument has the form
where ? is the symbol that denotes therefore.
The statement ((p ? q) p) ? q is a tautology.

p ? q p ? q
4
Rules of Inference Rule for Propositional Logic

The argument form with premises p1, p2, ., pn
and conclusion q is valid, when (p1p2.pn) ? q
is a tautology.
To show that an argument is valid, instead of
showing by truth table, we can establish the
validity of some relatively simple argument
forms, called rules of inference, which can be
used as building blocks to construct more
complicated valid argument forms.
The tautology ((p ? q) p) ? q is the basis of
the rule of inference called modus ponens or law
of detachment.

5
Rules of Inference Rules for Propositional Logic

Example Suppose that the conditional statement
If it snows today, then we will go skiing and
its hypothesis It is snowing today, are true.
Then by modus ponens, it follows that the
conclusion of the conditional statement, We will
go skiing is true.
Q1 Determine whether the argument given here is
valid and determine whether its conclusion must
be true because of the validity of the argument.

6
Rules of Inference Rules for Propositional Logic
7
Rules of Inference Rule for Propositional Logic

Example State which rule of inference is the
basis of the following argument It is below
freezing now. Therefore, it is either below
freezing or raining now.
Sol Let p It is below freezing now. and q
It is raining now. Then this argument is of the
form
p
?p v q
This is an argument that uses the addition rule.
Q2 State which rule of inference is the basis of
the following argument It is below freezing and
raining now. Therefore, it is below freezing.

8
Introduction to Proofs

A theorem is a statement that can be shown to be
true (usually important statement)
Less important theorem sometimes are called
propositions
A proof is a sequence of statements (valid
argument) to show that a theorem is true
The statements to be used in proofs include
Axioms (statement assumed to be true without
proof)
Ex If x is positive integer then x1 is positive
integer.
Hypothesis (premises) of the theorem
Previously proven theorems
Rules of inference used to draw conclusions and
to move from one step to another

9
Introduction to Proofs
Axioms
Rules of inference
New theorem
Hypothesis
proven theorems

A less important theorem that is helpful in the
proof of other results is called a lemma
A corollary is a theorem that can be established
directly from a theorem that has been proved
A conjuncture is a statement that is being
proposed to be a true statement, usually on the
basis of some partial evidence
When a proof of a conjuncture is found, the
conjuncture becomes a theorem

10
Introduction to Proofs

Example 1 If I have a car (C) I will drive to
Makkah (M). My boss gave me 60,000 (G) or Fired
me (F). If I have SR60,000 (H) then I have a car
(C). My boss did not fire me. Therefore I will
drive to Makkah (M).
G?F Hypothesis
?F Hypothesis
G Disjunctive syllogism rule using 1 and
2
G?H Axiom ?
H ?C Hypothesis
G ?C Hypo. syllogism using 4,5
C Modus ponens using 3 and 6
C ?M Hypothesis
M Modus ponens using 7 and 8

11
Methods of proving theoremsDirect proofs

A direct proof of a conditional statement p?q is
constructed when the first step is the assumption
that p is true subsequent steps are constructed
using rules of inference, with the final step
showing that q must also be true
In a direct proof, we assume that p is true and
use axioms, definitions, and previously proven
theorems, together with rules of inference, to
show that q must also be true
Def The integer n is even if there exists an
integer k such that n 2k, and n is odd if there
exists an integer k such that n 2k 1.

12
Methods of proving theoremsDirect proofs

Example 2 Use a direct proof to show that if n
is even then n2 is even
Proof Assume that n is even (hypothesis)
gt n 2k where k is integer (definition of even
number)
gt n2 (2k)2 4k2 2(2k2) (By squaring)
Since r 2K2 is integer (Axiom)
gt n2 2r is even
Q 3 Use a direct proof to show that if n is odd
then n2 is odd

13
Methods of proving theoremsProof by
contraposition

An indirect proof of a conditional statement p ?q
is a direct proof of its contraposition ?q ? ?p.
Example 3 Use an indirect proof to show that if
a and b are integers, and (a b) 15, then a
8 or b 8.
Proof The contraposition of (a b 15) ? (a
8) v (b 8) is (a lt 8) ? (b lt 8) ? (a b lt 15)
Suppose (a lt 8) ? (b lt 8) (hypothesis).
gt (a 7) ? (b 7),
gt (a b) 14,
gt (a b) lt 15.

14
Methods of proving theoremsProof by
contraposition

Example 4 Use an indirect proof to show that if
n2 is even then n is even
Proof The contraposition is if n is not even
then n2 is not even
Assume that n is not even i.e., n is odd
(hypothesis)
gt n 2k1 where k is integer (definition of
odd number)
gt n2 (2k 1)2
4k2 4k 1
2(2k2 2k) 1 2r 1, where r 2k2
2k
Since r is integer (Axiom) gt n2 is not even
Q 4 Use an indirect proof to show that if n is
odd then n2 is odd

15
Methods of proving theoremsProof by
contraposition

Example 5 Prove that if n ab then a ? ?n or b
? ?n where a and b are positive integers
Proof Let pa??n, qb??n and rnab
We want to prove that r? p?q
gt The contraposition is ?(p?q ) ? ? r (By
definition)
gt ?p ? ?q ? ? r (De Morgans law)
Now, assume that a??n and b??n (?p ? ?q)
gt a.b ? ?n.?n n (by multiplying above twos)
gt ab ? n
gt ? r

16
Methods of proving theoremsVacuous Proofs

Vacuous Proofs A conditional
statement p ? q is TRUE
if p is FALSE. If we can show
that p is False, then we have
a proof, called vacuous proof,
of the conditional statement p ? q
Example 6 Prove that if x2 ? 0 then 12 where x
is a real number
Proof Since x2 ? 0 for every real number then
the implication is vacuously true
Example 7 Prove that if he is alive and he is
dead then the sun is ice cold.
Proof Since the hypothesis is always false the
implication is vacuously true.

p q p ? q
F F T T F T F T T T F T
17
Methods of proving theoremsTrivial Proofs

Trivial Proofs A conditional
statement p ? q is TRUE if q
is TRUE. If we can show that
q is TRUE, then we have a
proof, called trivial proof, of
the conditional statement p ? q
Example 8 Prove that if x2 then x2 ? 0 for all
real numbers
Proof Since x2 ? 0 is true then the implication
is trivially true. (we didnt use the fact x2)
Q 5 Use a trivial proof to show that if n gt 1
then n2 n for all integers

p q p ? q
F F T T F T F T T T F T
18
Methods of proving theoremsProofs by
Contradiction

Proof by Contradiction To prove a proposition p,
assume not p and show a contradiction.
Example 9 Use a proof by contradiction to show
that ?2 is irrational
Proof Let ?2 is rational
gt ?2 a/b for some integers a and b (b?0)
(relatively prime).
(Definition of rational numbers)
gt 2 a2/b2 (Squaring both sides)
gt 2b2 a2
gt a2 is even (Definition of even numbers)
gt a is even (a 2k for some k)
gt 2b2 a2 (2k)2 4k2
gt b is even (Definition of even numbers)
But if a and b are both even, then they are not
relatively prime!
Hence, ?2 is irrational

19
Methods of proving theoremsProofs by
Contradiction

Why is this method valid ?
The contradiction forces us to reject our
assumption because our other steps based on that
assumption are logical and justified. The only
mistake that we could have made was the
assumption itself.
Be careful!
Sometimes the contradiction comes from a mistake
in the steps of the proof and not from the
assumption. This makes the proof invalid.
Example 10 Prove that 12
Proof Suppose that 2?1 and ab for some a.
gt 2b ?b multiply by b
gt ab ?b 2bbbab by hypothesis
gt (a-b)(ab) ? b(a-b) multiply by a-b
gt a2-b2 ? ab-b2
gt a2 ?ab subtract b2
from both sides
gt a ?b which contradicts our assumption that
ab
Hence it follows that 12
Can you find the error ?

20
Methods of proving theoremsProofs by
Contradiction

To prove a conditional statement p ? q by
contradiction we prove that p ? ?q? F is true
which is equivalent to p ? q .
Example 11 Use a proof by contradiction to show
that If 3n2 is odd then n is odd
Proof Suppose that 3n2 is odd and n is even p
? ?q
gt n 2r hypothesis ?q, definition of
even numbers
gt 3n 6r multiply 1 by 3
gt 3n226r add 2 to both sides
gt 3n22(13r)
gt 3n22k let k13r
gt Thus 3n2 is even which is false (a
contradiction !)
Therefore the implication is true.

21
Methods of proving theoremsProofs of Equivalence

Proofs of Equivalence To prove p ?q we have to
prove p ? q and q ? p
Example 12 prove n is even if and only if n2 is
even
if n is even then n2 is even proved in example
2
n2 is even then n is even proved in example 4
Therefore, n is even if and only if n2 is even
Proving equivalence of several propositions If
we want to prove that p1? p2 ? p3 ? pn
Then it is sufficient to prove p1 ? p2,, p2 ? p3
pn ? p1
Disproof by Counterexample
Example 13 Prove that For all real numbers x2 gt
x is false
Proof X0.5 is a counterexample since 0.52 gt 0.5
is not true
Q 6 Prove that If n is not positive, then n2 is
not positive is false

22
Methods of proving theorems Proof by Cases

Proof by Cases A proof by cases must cover all
possible cases that arise in a theorem. Each case
may cover an infinite number of instances
Example 15 Prove "if n is an integer then n2
n".
Proof We can prove that n2 n for every integer
by considering three cases, when n 0, when n
1, and when n -1.
Case 1 When n 0, since 02 0, 02 gt 0. So,
n2 n is true.
Case 2 When n 1, From n 1, we get n2 n
(by multiplying with n).
Case 3 When n -1, since n is negative, n2 is
positive, so n2 n.
Hence, if n is an integer then n2 n.
Q 7 Prove that xy xy, where x and y real
numbers.
Hint Consider all 4 cases x and y positive or
negative
Common errors with exhaustive proof and proof by
cases
Draw conclusion from non-exhaustive examples
Not covering all possible cases

23
Methods of proving theorems Existence Proofs

Many theorems state that an object with certain
properties exists, i.e., ?xP(x) where P is a
predicate. A proof of such a theorem is called an
existence proof. There are two kinds
Constructive Existence Proof The proof is
established be giving example a such that P(a) is
true
Example 16 Prove "there is a positive integer
that can be written as the sum of cubes in two
different ways.
Proof Consider 17291039312313. Finding such
examples may require computer assistance.
Non-constructive Existence Proof The proof is
established by showing that an object a with P(a)
is true must exist without explicitly
demonstrating one. Proofs by contradiction are
usually used in such cases.

24
Methods of proving theorems Existence Proofs

Example 17 Let x1,x2,..,xn be positive integers
such that their average is m. prove that there
exists xi such that xi m
Proof Suppose that there is no such number,
i,e.,
x1 ? m, x2 ? m, , xn ? m
By adding these inequalities we get x1 x2
xn ? nm
Dividing by n (x1 x2 xn)/n ? m
But since the average is defined as (x1 x2
xn)/n
Then we have m ? m which is a contradiction.
Therefore there must be a number xi such that xi
m. But we can not specify which number is that.

25
Methods of proving theorems Uniqueness Proofs

Some theorems state that there is exactly one
element with a certain property. A proof of such
a theorem is called a uniqueness proof.
Strategy here is (1) show that an element x with
the desired property exists (2) show that any
other y (y ! x) does not have the property,
i.e., if x and y both have the property, then x
must equal y.
Example 18 Prove that the equation 3x5 9 has
a unique solution.
Proof (1) There exists a solution namely x 4/3
(2) Suppose that y and z are solution then
3y5 9 3z5
So 3y 3z
Dividing by 3 we get y z
This proves that the solution is unique

26
Methods of proving theoremsMathematical
Induction

Principle of Mathematical Induction
Let P(n) be a statement for all the positive
integers (n 1, 2, 3, . .). If the following two
properties hold
P(1) is true.
P(k1) is true if P(k) is true for each positive
integer k.
Then P(n) is true for all n.
First part is a simple proposition we call the
base step
Second part is an inductive step. Start by
assuming P(k) is true, and show that P(k1) is
also true
The assumption that P(k) is true called the
inductive hypothesis
So, we prove that
(P(1) ? ?k (P(k) ? P(k1))) ? (?n P(n))

27
Methods of proving theoremsMathematical
Induction

Example 19 Suppose we have an infinite ladder,
and we want to know whether we can reach every
step on this ladder. We know two things
We can reach the first rung of the ladder
If we can reach a particular rung of the ladder,
then we can reach the next rung.
How does Induction Work?
Consider the above example. The rules for
reaching steps can help you remember how
induction works
Statements (1) and (2) are the basic step and
inductive step respectively of the proof that
P(n) is true for all positive integers n, where
P(n) is the statement that we can reach the nth
rung of the ladder.
Consequently, we can invoke the mathematical
induction to conclude that we can reach every
rung of the ladder

28
Methods of proving theoremsMathematical
Induction

Example 20 (A Summation Problem) Prove that for
any integer n 1 1 2 3 n
n(n1)/2 .
Proof Let P(n) be the proposition that the sum
of the first n positive integers is n(n1)/2.
Then to proof that P(n) is true for all n 1, we
have to show that P(1) is true and P(k1) is true
if P(k) is true for k 1.
Basic step P(1) is true, because 1 1.(11)/2
Inductive step Let us assume that it is true
for n k, that is, 123..k k(k1)/2
Now, if we can prove that it is true for n k1
also, then it can be said that it is true for all
n.
For, 1 2 k (k 1)
k(k 1)/2 (k 1)
(k 1)(k/2 1)
(k 1)(k 2)/2
gt P(k 1) is also true, hence P(n) is true for
all integer n.

29
Methods of proving theoremsMathematical
Induction

Example 21 Use induction to prove that the sum
of the first n odd integers is n2.
Proof Let P(n) be the proposition that the sum
of the first n odd integers is n2. Then to proof
that P(n) is true for all n 1, we have to show
that P(1) is true and P(k1) is true if P(k) is
true for k 1.
Basic step P(1) is true, because the sum of the
first 1 odd integer is 12.
Inductive step Assume P(k) the sum of the
first k odd integers is k2, that is, 1 3
(2k - 1) k2
Now, if we can prove that it is true for n k1
also, then it can be said that it is true for all
n.
For, 1 3 (2k-1) (2k1)
k2 (2k 1)
(k1)2
gt P(k 1) is also true, hence P(n) is true for
all integer n.

30
Methods of proving theoremsMathematical
Induction

Q 8 Use induction to prove that
1?1! 2?2! n?n! (n1)! - 1, ?n
Q 9 Use induction to prove that for all n,
Q 10 Use induction to prove that for all n,

31
Methods of proving theoremsMathematical
Induction

Example 22 Use induction to prove the inequality
n lt 2n, ?n gt 0.
Proof Let P(n) be the proposition that n lt 2n
Basic step P(1) is true, because 1 lt 21 2.
Inductive step Assume P(k) is true, that is, k
lt 2K
Now, if we can prove that it is true for n k1
also, then it can be said that it is true for all
n.
For, k lt 2K
gt k 1 lt 2K 1
2K 2K
2. 2K
2K1
gt k 1 lt 2K1
gt P(k 1) is also true, hence P(n) is true for
all integer n.
Q 11 Use induction to prove the inequality 2n lt
n! ?n gt 3.

32
Methods of proving theorems Recursive
Definitions

Recursion Sometimes it is difficult to define an
object explicitly. However, it may be easy to
define this object in terms of itself. This
process is called recursion.
Recursive defined functions We use two steps to
define a function with the nonnegative integers
as its domain
Basis Step Specify the value of the function at
zero.
Recursive Step Give a rule for finding its
value at an integer from its values at smaller
integers.
Such definition is called a recursive or
inductive definition.
Example 25 The definition of factorial function
n! 1 2 3 (n-1) n, n ? 1
But equivalently, we could define it like this

33
Methods of proving theorems Recursive
Definitions

Example 26 The definition of Fibonacci Numbers

34
Methods of proving theorems Recursive
Definitions

Example 27 Suppose that f is defined by
f(0) 3,
f(n1) 2f(n) 3.
Find f(1), f(2), f(3), and f(4).
Solution From the recursive definition, it
follows that
f(1) 2f(0) 3 2.3 3 9,
f(2) 2f(1) 3 2.9 3 21,
f(3) 2f(2) 3 2.21 3 45,
f(4) 2f(3) 3 2.45 3 93.
Example 28 Give a recursive definition of
Solution The first and second part of the
recursive definition are

35
Methods of proving theorems Recursive
Definitions

Example 29 Give an inductive definition of S
x x is a multiple of 3
Solution
3 ? S
x,y ? S ? x y ? S
x,y ? S ? x - y ? S
No other numbers are in S.
Q 12 Find the Fibonacci numbers f(2), f(3),
f(4), f(5), and f(6).
Q 13 Give an inductive definition of an.

36
Methods of proving theorems Recursive
Definitions

Recursively Defined Sets and Structures Sets can
be defined recursively. Recursive definition of
sets have two part basis step and recursive
step.
Basis Step An initial collection of elements is
specified.
Recursive Step Rules for forming new elements
in the set from those already known to be in the
set are provided.
Example 30 Consider subset S of the set of
integers defined by
Basis Step 3? S.
Recursive Step If x, y ? S then x y ? S .
The new elements found to be in S are 3 by the
basis step,
336 at the first application of the recursive
step,
36639 at the second application of the
recursive step,
6612 at the third application of the recursive
step, and so on.
We will show later that S is set of all positive
multiples of 3.

37
Recursive DefinitionsRecursively Defined Sets

Example 31 The set of Natural Numbers N can be
defined recursively as follows
1?N basis
step
If x ?N then x1?N Recursive step
Lets try to constructs the set using this
definition
1 in N basis step
112 in N Recursive step
213 in N Recursive step
Etc

Algorithms

39
AlgorithmsDefinitions

The word algorithm comes from the name of a
Persian author, Abu Jafar Mohammad Ibn Musa Al
Khowarizmi (825 AD).
Definition An algorithm is a finite set of
instructions that, if followed, carries out a
particular task. In addition, all algorithms must
satisfy the following criteria
Input Zero or more quantities are externally
supplied.
Output At least one quantity is produced.
Definiteness Each instruction is clear and
unambiguous.
Finiteness If we trace out the instructions of
an algorithm, then for all cases, the algorithm
terminates after a finite number of steps.
Effectiveness Every instruction must be very
basic so that it can be carried out, in
principle, by a person using only pen and paper.
It must be feasible.

40
Study of Algorithms

How to devise algorithms?
Creating an algorithm is an art which may never
be fully automated.
How to validate algorithms?
Once an algorithm is devised, it is necessary to
show that it computes the correct answer for all
possible legal inputs.
A program can be written and then be verified.
How to test a program?
Debugging is the process of executing programs on
sample data sets to determine whether the faulty
results occur and, if so, to correct them.
Profiling is the process of executing a correct
program on data sets and measuring the time and
space it takes to compute the results.

41
Pseudocode Conventions

Comments begin with // and continue until the end
of line.
Blocks are indicated with matching braces and.
An Identifier begins with a letter max.
Assignment of values to variables is done using
assignment statement Variableexpression.
Logical operators and, or and not are provided.
Relational operators lt, , , ?, and gt
provided.
Elements of arrays are accessed using and .
While loop
while (condition) do
statements

42
Pseudocode Conventions

For loop
for variablevalue1 to value2 step
step do
Statements
Repeat-until loop
repeat
Statements
until (condition)
Conditional statement
if (condition) then (statement)
if (condition) then (statement 1)
else (statement 2)

43
Pseudocode Conventions

Case statement
case
(condition 1) (statement 1)
.
(condition n) (statement n)
else (statement n1)
Input and output are done using read and write.
There is only one type of procedure Algorithm.
An algorithm consists of a heading and a body.
The heading of an algorithm takes the form
Algorithm Name ((parameter list))

44
Finding Maximum Value

Input A sequence of n numbers (a1, a2,, an).
Output Maximum of (a1, a2,, an).
Algorithm Maximum(A, n)
MaxA1
for i2 to n do
if ( Max lt Ai) then
MaxAi

45
Sorting Problem

Input A sequence of n numbers (a1, a2,, an).
Output A permutation of n numbers (reordering)
(a'1, a'2,, a'n) of the input sequence such that
a'1 a'2 a'n.
Bubble Sort-
It is a popular sorting algorithm
It swaps adjacent elements that are out of order
Insertion sort-
Insert an element to a sorted array such that the
order of the resultant array be not changed.

46
Bubble Sort Algorithm