MIT School of Computing
Department of Computer Science & Engineering
Department of Computer Science & Engineering
21BTAS305 -Discrete Mathematics
PLD
Applicable for Class - S.Y(1-18) (SEM-I)
AY 2023-2024 SEM-I
Designed By Team : Discrete Mathematics
MIT School of Computing
Department of Computer Science & Engineering
Course Outcomes:
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1. Ability to apply set theory and logic for problem solving and
inferencing.
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2. Ability to solve problems using Permutations, combinations,
probabilities, relations and functions.
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PLD
3. Ability to apply the concepts of graphs and trees for problem solving.
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ByTeam
Team :: Discrete
Mathematics
Designed
Discrete
Mathematics
MIT School of Computing
Department of Computer Science & Engineering
1. SETS AND PROPOSITIONS
Sets, Combination of sets, Finite and Infinite sets, Uncountable infinite sets, Principle of
inclusion and exclusion, multi-sets, Mathematical Induction. Propositions, Conditional
Propositions, Logical Connectivity, Propositional calculus, Universal and Existential
Quantifiers, Normal forms, Rules of inference, Predicate calculus, methods of proofs.
2. PERMUTATIONS, COMBINATIONS ANDPLD
DISCRETE PROBABILITY
Permutations and Combinations: rule of sum and product, Permutations, Combinations,
Algorithms for generation of Permutations and Combinations, binomial theorem,
Discrete Probability, Conditional Probability, Bayes’ Theorem, Information and Mutual
Information.
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ByTeam
Team :: Discrete
Mathematics
Designed
Discrete
Mathematics
COURSE CONTENTS
3. RELATIONS AND FUNCTIONS
A relational model for data bases, Properties of Binary Relations, Closure
of relations, Warshall’s algorithm, Equivalence relations and partitions,
Partial ordering relations and lattices, Chains and Anti chains, Compatible
relations. Functions, Composition of functions, Invertible functions,
Pigeonhole Principle, Recursive function.
4. GRAPH THEORY
Basic terminology, multi-graphs and weighted graphs, representation of
graphs, Subgraphs, Isomorphic graphs, Complete, regular and bipartite
graphs, operations on graph, paths and circuits, graph traversals,
Hamiltonian and Euler paths and circuits, shortest path in weighted
graphs (Dijkstra’s algorithm), factors of a graph, planer graph and
Traveling salesman problem, Graph Coloring.
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COURSE CONTENTS
5. TREES AND CUT SETS
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Basic terminology and characterization of trees, rooted
path lengths in rooted trees, Prefix codes and optimal
codes, binary search trees, Tree traversal, Spanning
Fundamental Trees and cut sets, Minimal Spanning
Kruskal’s and Prim’s algorithms for minimal spanning
The Max flow-Min Cut Theorem (Transport network).
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Note: A student group should make one Mathematical Model
and submit it as a part of Tutorial.
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trees,
prefix
trees,
trees,
trees,
Unit 1: SETS AND PROPOSITIONS
Sets, Combination of sets, Finite and Infinite sets,
Uncountable infinite sets, Principle of inclusion and
exclusion,
multi-sets,
Mathematical
Induction.
Propositions,
Conditional
Propositions,
Logical
Connectivity, Propositional calculus, Universal and
Existential Quantifiers, Normal forms, Rules of inference,
Predicate calculus, methods of proofs.
Designed By Team : Discrete Mathematics
Introduction to Set Theory
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A set is a structure, representing an unordered
collection (group, plurality) of zero or more distinct
(different) objects.
Well / proper defined collection of objects.
Set theory deals with operations between, relations
among, and statements about sets.
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Basic notations for sets
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For set names, we’ll use capital letters.
For set elements / objects, we’ll use small letters.
We can denote a set S in writing by listing all of its
elements in curly braces:
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{a, b, c} is the set of whatever 3 objects are denoted by a, b,
c.
Set builder notation: For any proposition P(x) over
any universe of discourse, {x |P(x)} is the set of all x
such that P(x).
e.g., {x | x is an integer where x>0 and x<5 }
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Basic properties of sets
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Sets are inherently unordered:
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No matter what objects a, b, and c denote,
{a, b, c} = {a, c, b} = {b, a, c} =
{b, c, a} = {c, a, b} = {c, b, a}.
All elements are distinct (unequal);
multiple listings make no difference!
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{a, b, c} = {a, a, b, a, b, c, c, c, c}.
This set contains at most 3 elements!
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Definition of Set Equality
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Two sets are declared to be equal if and only if
they contain exactly the same elements.
In particular, it does not matter how the set is
defined or denoted.
For example: The set {1, 2, 3, 4} =
{x | x is an integer where x>0 and x<5 } =
{x | x is a positive integer whose square
is >0 and <25}
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Infinite Sets
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Conceptually, sets may be infinite (i.e., not finite,
without end, unending).
Symbols for some special infinite sets:
N = {0, 1, 2, …} The natural numbers.
Z = {…, -2, -1, 0, 1, 2, …} The integers.
R = The “real” numbers, such as
374.1828471929498181917281943125…
Infinite sets come in different sizes!
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Some Predefined Set
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N = { 0, 1, 2, 3, … } is the set of natural numbers.
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Z = { …, -2, -1, 0, 1, 2, … } is the set of integers.
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Z+ = { 1, 2, 3, … } is the set of positive integers.
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Q = { p/q | p, q  Z and q  0 } is the set of Rationals.
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R = the set of real numbers.
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Basic Set Relations: Member of
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xS (“x is in S”) is the proposition that object x is
an lement or member of set S.
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e.g. 3N, “a”{x | x is a letter of the alphabet}
Can define set equality in terms of  relation:
S,T: S=T  (x: xS  xT)
“Two sets are equal iff they have all the same
members.”
xS : (xS) “x is not in S”
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The Empty Set
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 (“null”, “the empty set”) is the unique set that
contains no elements whatsoever.
 = {} = {x|False}
No matter the domain of discourse,
we have the axiom
x: x.
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Subset and Superset Relations
ST (“S is a subset of T”) means that every
element of S is also an element of T.
• ST  x (xS  xT)
• S, SS.
• ST (“S is a superset of T”) means TS.
• Note S=T  ST ST.
• S 
/ T means (ST), i.e. x(xS  xT)
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Proper (Strict) Subsets & Supersets
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ST (“S is a proper subset of T”) means that ST
but Similar for ST.
T / S
Example:
{1,2} 
{1,2,3}
S
T
Venn Diagram equivalent of ST
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Sets Are Objects, Too!
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The objects that are elements of a set may
themselves be sets.
E.g. let S={x | x  {1,2,3}}
then S={,
{1}, {2}, {3},
{1,2}, {1,3}, {2,3},
{1,2,3}}
Note that 1  {1}  {{1}} !!!!
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Cardinality and Finiteness
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|S| (read “the cardinality of S”) is a measure of
how many different elements S has.
E.g., ||=0, |{1,2,3}| = 3, |{a,b}| = 2,
|{{1,2,3},{4,5}}| = ____
We say S is infinite if it is not finite.
What are some infinite sets we’ve seen?
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The Power Set Operation
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The power set P(S) of a set S is the set of all
subsets of S. P(S) = {x | xS}.
E.g. P({a,b}) = {, {a}, {b}, {a,b}}.
Sometimes P(S) is written 2S.
Note that for finite S, |P(S)| = 2|S|.
It turns out that |P(N)| > |N|.
There are different sizes of infinite sets!
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Ordered n-tuples
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For nN, an ordered n-tuple or a sequence of
length n is written (a1, a2, …, an). The first
element is a1, etc.
These are like sets, except that duplicates matter,
and the order makes a difference.
Note (1, 2)  (2, 1)  (2, 1, 1).
Empty sequence, singlets, pairs, triples,
quadruples, quintuples, …, n-tuples.
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Cartesian Products of Sets
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For sets A, B, their Cartesian product
AB : {(a, b) | aA  bB }.
E.g. {a,b}{1,2} = {(a,1),(a,2),(b,1),(b,2)}
Note that for finite A, B, |AB|=|A||B|.
Note that the Cartesian product is not
commutative: AB: AB =BA.
Extends to A1  A2  …  An...
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The Union Operator
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For sets A, B, their union AB is the set
containing all elements that are either in A, or
(“”) in B (or, of course, in both).
Formally, A,B: AB = {x | xA  xB}.
Note that AB contains all the elements of A and
it contains all the elements of B:
A, B: (AB  A)  (AB  B)
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Union Examples
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{a,b,c}{2,3} = {a,b,c,2,3}
{2,3,5}{3,5,7} = {2,3,5,3,5,7} ={2,3,5,7}
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The Intersection Operator
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For sets A, B, their intersection AB is the set
containing all elements that are simultaneously in
A and (“”) in B.
Formally, A,B: AB{x | xA  xB}.
Note that AB is a subset of A and it is a subset
of B:
A, B: (AB  A)  (AB  B)
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Intersection Examples
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
{a,b,c}{2,3} = ___
{4}
{2,4,6}{3,4,5} = ______
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Disjointedness
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Two sets A, B are called
disjoint (i.e., unjoined)
iff their intersection is
empty. (AB=)
Example: the set of even
integers is disjoint with
the set of odd integers.
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Help, I’ve
been
disjointed!
Set Difference
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For sets A, B, the difference of A and B, written
AB, is the set of all elements that are in A but
not B.
A  B : x  xA  xB
 x   xA  xB  
Also called:
The complement of B with respect to A.
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Set Difference Examples
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{1,2,3,4,5,6}  {2,3,5,7,9,11} = {1,4,6}
___________
Z  N  {… , -1, 0, 1, 2, … }  {0, 1, … }
= {x | x is an integer but not a nat. #}
= {x | x is a negative integer}
= {… , -3, -2, -1}
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Set Difference - Venn Diagram
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A-B is what’s left after B
“takes a bite out of A”
Chomp!
Set
AB
Set A
Set B
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Set Complements
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The universe of discourse can itself be considered
a set, call it U.
The complement of A, written A, is the
complement of A w.r.t. U, i.e., it is UA.
E.g., If U=N,
{3,5}  {0,1,2,4,6,7,...}
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More on Set Complements
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An equivalent definition, when U is clear:
A  {x | x  A}
A
A
U
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Set Identities
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Identity:
A=A AU=A
Domination: AU=U A=
Idempotent: AA = A = AA
Double complement: ( A )  A
Commutative: AB=BA AB=BA
Associative: A(BC)=(AB)C
A(BC)=(AB)C
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DeMorgan’s Law for Sets
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Exactly analogous to (and derivable from)
DeMorgan’s Law for propositions.
A B  A  B
A B  A  B
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Proving Set Identities
To prove statements about sets, of the form
E1 = E2 (where Es are set expressions), here are
three useful techniques:
• Prove E1  E2 and E2  E1 separately.
• Use logical equivalences.
• Use a membership table.
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Method 1: Mutual subsets
Example: Show A(BC)=(AB)(AC).
• Show A(BC)(AB)(AC).
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Assume xA(BC), & show x(AB)(AC).
We know that xA, and either xB or xC.
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Case 1: xB. Then xAB, so x(AB)(AC).
Case 2: xC. Then xAC , so x(AB)(AC).
Therefore, x(AB)(AC).
Therefore, A(BC)(AB)(AC).
Show (AB)(AC)  A(BC). …
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Method 2: Membership Tables
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Just like truth tables for propositional logic.
Columns for different set expressions.
Rows for all combinations of memberships in
constituent sets.
Use “1” to indicate membership in the derived set,
“0” for non-membership.
Prove equivalence with identical columns.
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Membership Table Example
Prove (AB)B = AB.
A
0
0
1
1
B AB (AB)B AB
0
0
0
0
1
1
0
0
0
1
1
1
1
1
0
0
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Membership Table Exercise
Prove (AB)C = (AC)(BC).
A
0
0
0
0
1
1
1
1
B
0
0
1
1
0
0
1
1
C A B ( AB )  C A C
0
1
0
1
0
1
0
1
B C
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(AC)(BC)
Inclusion-Exclusion Principle
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How many elements are in AB?
|AB| = |A|  |B|  |AB|
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|AB C | = |A|  |B|  |C|  |AB| 
|BC|  |AC|+  |AB C |
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Inclusion-Exclusion Principle
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In a town of 10000 families it was found that
40% of families buy newspaper A, 20% family
buy newspaper B, 10% family buy newspaper
C, 5% family buy newspaper A and B, 3%
family buy newspaper B and C and 4% family
buy newspaper A and C. If 2% family buy all
the newspaper. Find the number of families
which buy
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1.
2.
3.
4.
5.
6.
Number of families which buy all three newspapers.
Number of families which buy newspaper A only
Number of families which buy newspaper B only
Number of families which buy newspaper C only
Number of families which buy exactly only one
newspaper
Number of families which buy None of A, B, C
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Designed By Team : Discrete Mathematics
Designed By Team : Discrete Mathematics
Designed By Team : Discrete Mathematics
Mathematical Induction:
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Mathematical Induction is a mathematical technique which is used to
prove a statement, a formula or a theorem is true for every natural
number.
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The technique involves two steps to prove a statement, as stated
below −
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Step 1(Base step) − It proves that a statement is true for the initial
value.
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Step 2(Inductive / induction step) − As step 1 is true so put n=k and
consider it true, and put n= k+1 in given expression.
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Mathematical Induction:
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Eg:
1+2+3+…….+n=n(n+1)/2 where n>=1 prove this
expression by mathematical induction.
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1+2+22+ ….. + 2n = 2n+1 – 1 prove using
mathematical induction principal where n >=0
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2+4+6 ….. + 2n = n(n+1) prove using mathematical
induction principal where n >=1
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Propositions
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A statement or expression which is either true or false
but not both.
Some valid Eg:
Invalid Proposition
2+2=5
The earth is flat.
X+3=5
Bring the book.
There are 7 days in week. When is your interview?
Will there be holiday
tomorrow?
What a beautiful painting.
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Propositions
True value
is denoted
by T or 1
False value
is denoted
by F or 0
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Propositions :Connectives
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In propositional logic generally we use five connectives
which are −
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OR (∨) : Disjunction
AND (∧) : Conjunction
Negation / NOT (¬)
Implication / if-then (→) conditional
If and only if (⇔) biconditional.
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Propositions :Connectives
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If and only if (⇔) − A⇔ is bi-conditional logical
connective which is true when p and q are same, i.e.
both are false or both are true.
The truth table is as follows −
A
B
A⇔B
True
True
True
True
False
False
False
True
False
False
False
True
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Propositions :Connectives
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OR (∨) − The OR operation of two propositions A and B
(written as A∨B) is true if at least any of the
propositional variable A or B is true.
The truth table is as follows −
A
B
A∨B
True
True
True
True
False
True
False
True
True
False
False
False
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Propositions :Connectives
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AND (∧) − The AND operation of two propositions A
and B (written as A∧B) is true if both the propositional
variable A and B is true.
The truth table is as follows −
A
B
A∧B
True
True
True
True False False
False True False
False False False
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Propositions :Connectives
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Negation (¬) − The negation of a proposition A (written
as ¬A) is false when A is true and is true when A is false.
The truth table is as follows −
A
¬A
True
False
False
True
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Propositions :Connectives
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Implication / if-then (→) − An implication A→B is the
proposition “if A, then B”. It is false if A is true and B is
false. The rest cases are true.
The truth table is as follows −
A
B
A→B
True
True
True
True
False
False
False
True
True
False False
True
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Propositions :Connectives
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Product: This is nothing but same like biconditional.
The product is represented by symbol x. its value is
true when both the expressions are equal.
A
B
AxB
True
True
True
True
False
False
False
True
False
False
False
True
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A
B
A⇔B
A
B
A→B
True
True
True
True
True
True
True
False
False
True
False
False
False
True
False
False
True
True
False
False
True
False
False
True
A
B
A∨B
A
B
A∧B
True
True
True
True
True
True
True
False
True
True
False
False
False
True
True
False
True
False
False False
False
False False False
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Propositional Decisions
A Tautology is a formula which is
always true for every value of its
propositional variables.
A Contradiction is a formula which
is always false for every value of
its propositional variables.
A Contingency is a formula which
has both some true and some
false values for every value of its
propositional variables.
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Propositional Decisions
Example − Prove [(A→B)∧A]→B is a tautology
The truth table is as follows −
A
B
A→B
(A → B) ∧ A
[( A → B ) ∧ A] → B
True
True
True
True
True
True
False
False
False
True
False
True
True
False
True
False
False
True
False
True
As we can see every value of [(A→B)∧A]→B is "True", it is a
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tautology.
Propositional Decisions
Example − Prove (A∨B)∧[(¬A)∧(¬B)] is a contradiction
The truth table is as follows −
.
A
B
A∨B
¬A
¬B
(¬ A) ∧ ( ¬ B)
(A ∨ B) ∧ [( ¬ A) ∧
(¬ B)]
True
True
True
False
False
False
False
True
False
True
False
True
False
False
False
True
True
True
False
False
False
False
False
False
True
True
True
False
As we can see every value of (A∨B)∧[(¬A)∧(¬B)] is “False”, it is
a contradiction
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Propositional Decisions
Example − Prove (A∨B)∧(¬A) a contingency
The truth table is as follows −
A
B
A∨B
¬A
(A ∨ B) ∧ (¬
A)
True
True
True
False
False
True
False
True
False
False
False
True
True
True
True
False
False
False
True
False
As we can see every value of (A∨B)∧(¬A) has both “True” and
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“False”, it is a contingency.
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Prove following expressions are logically
equivalent using truth table.
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1. p-> q = ~ p v q
2. p <-> q = (p->q) ∧ (q->p)
3. p-> (q->r) = (p∧~r)->~q
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Normal Forms
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One of the major problem of logical expression is that
to find whether the given statement is tautology or
contradiction using truth table so it can be done by
another way to desired result considering expression
only in conjunction and disjunction using logical
identities is called as Normal forms.
Types:
DNF (Disjunctive Normal form)
CNF (Conjunctive Normal form)
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Normal Forms
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DNF (Disjunctive Normal form):
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Disjunction is placed in between the conjunction.
Or (∨) is placed in between the And (∧).
Eg
(~p ∧ q ) v (p ∧ q)
(p ∧ q) v ~p
(p ∧ ~q ) v (p ∧ q) v (~p ∧ q)
(p ∧ q ∧ r ) v ~r
p
->ByqTeam =
~p
vq
Designed
: Discrete
Mathematics
Normal Forms
•
CNF (Conjunctive Normal form):
•
•
•
•
•
•
Conjunction is placed in between the disjunction.
And (∧) is placed in between the Or (v).
Eg
(p ∧ q)
~p ∧ (p v q)
(p v q) ∧ (~p v ~q)
p
->ByqTeam =
~p
vq
Designed
: Discrete
Mathematics
Predicates
It is the verbal statement which describes the
property of a variable.
•
•
•
•
•
Usually represented by the letter
The notation P(x) is used to represent some
unspecified property or predicate that x may have.
– P(x) = x has 30 days.
– P(April) = April has 30 days.
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Predicates
One or more statements with some common
properties, we call that common property as
predicate.
• Eg: a(x) = x is cleaver.
x = Ram is cleaver.
Here x is called as place holder, after placing value in
place holder we can say that predicate become
proposition.
•
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Predicates
•
•
•
•
•
•
Predicates can have n number of placeholders.
Eg: (2 place holder)
A(x,y) = x is greater than y.
x= 7 , y= 5
7 is bigger than 5 (x >y), Q(x,y)
Here Q(x , y) shows the relation that x is greater than
y.
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Predicates
•
•
•
•
•
Eg: (3 place holders)
A(x,y,z) = y lies in between x and z .
x= 23 , y= 45 , z= 92
45 lies in between 23 and 92. Q(x,y,z)
Here Q(x , y , z) shows the relation that y lies in x and z.
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Predicates
•
•
•
Universe of Discourse: The values which the variables
may be assume constitute set called as to make
required condition true, is universe of discourse.
Eg: p(x,y): x+y=10 , Let universe be set of natural
numbers.
By putting x=1 we get one place predicate
p(1,y):1+y=10. further setting y=9 we obtain the
proposition p(1,9) which is true for given condition.
Designed By Team : Discrete Mathematics
Predicates
•
Universe of Discourse: set of all values which makes
the given predicate true.
•
Eg: p(x,y): x+y=10 , Let universe bet set of natural
numbers.
However if we set y=10, x=1 then p(1,10) is false
proposition.
so here we can say for all x and y the predicate can not
be true.
•
•
Designed By Team : Discrete Mathematics
The second
method of
binding the
variable in
predicate is by
quantification of
the variables.
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Quantifiers




•
•
It consist of special words which deals with
quantities such as some , all , few, etc.
It tells us about how much certain element is
present in given set.
It is an operator use to create the proposition from
propositional function .
Types:
Universal Quantifiers (∀ : for all)
Existential Quantifiers (∃ : there exist)
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Quantifiers
Universal
Quantifiers (∀ :
for all)
Existential
Quantifiers
The statement starts
with ∀ is called
universal quantifier &
statement ∀x f(x) is
called universally
quantified statement.
It consist of statement
that begins with some,
there exist, at least
one, etc.
It is denoted by ∃
∀(x)p(x) for every value
of x is true.
∃(x) p(x) for some
values of x is true.
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Quantifiers
•
∀(x) {M(x) -> H(x)
∀(x) {A(x) -> R(x)
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Quantifiers
•
∃(x){S(x) ∧ I(x)}
∃(x){A(x) ∧ H(x)}
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Existential Quantifier
•
•
•
Let P(x) be the predicate “x+3=5”
Then proposition ∃x p(x) is true if x=2
but ∀x p(x) is false.
Possible
Combination
s of two
place holder
predicate
using
quantifiers:
• ∃x ∀y p(x,y)
• ∀ y ∃x p(x,y)
• ∃x ∃y p(x,y)
• ∀ x ∀ y p(x,y)
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Existential Quantifier
•
•
•
•
•
Let , p(x,y) be the two place predicate, then
∃x ∀y p(x,y) is the proposition : There exist
value of x such that for all values of y p(x,y) is
true.
∀ y ∃x p(x,y) is the proposition: for each
value of y there exist x such that p(x,y) is
true.
∃x ∃y p(x,y) is the proposition: There exist
value of x & y such that p(x,y) is true.
∀ x ∀ y p(x,y) is the proposition: For all
values of x & y, p(x,y) is true.
Designed By Team : Discrete Mathematics
Negation of Quantified Statement
•
•
•
•
•
•
Consider example: All invited guest were present for
dinner: ∀x p(x).
The negation: All invited guest were not present for
dinner.
It means that some guest were not present for dinner. ∃x
[~p(x)] where,
x: x is a guest
P(x) : x was present for dinner.
∃x [~p(x)] and ∀x p(x) are equivalent.
Designed By Team : Discrete Mathematics
Negation of Quantified Statement
•
•
•
•
Consider example: There is student in class who is not
familiar with c programming. ∃x [~p(x)]
The negation: All students in class are familiar with c
programming.
Hence symbolically, ∃x [~p(x)] and ∀x p(x) are logically
equivalent.
∃x [~p(x)] and ∀x p(x) are equivalent.
Statement
Negation
∀x p(x)
∃x [~p(x)]
∃x [~p(x)]
∀x p(x)
∀x [~p(x)]
∃x [p(x)]
∃x [p(x)]Designed By Team : Discrete Mathematics
∀x [~p(x)]
Rules of Inference
(Deriving Conclusion from evidences)
Every Theorem in Mathematics,
or any subject for that matter, is
supported by underlying proofs.
These proofs are nothing but a
set of arguments that are
conclusive evidence of the validity
of the theory.
Designed By Team : Discrete Mathematics
Rules of Inference
(Deriving Conclusion from evidences)
•
•
•
•
Structure of an argument
An argument can be defined as a sequence of
statements we call these statements as premises
or assumptions.
The conclusion is used to indicate the last
statement, and premises are used to indicate all
the remaining statements.
Before the conclusion, the symbol ∴ will be
placed.
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Rules of Inference
Rules of Inference
(Deriving Conclusion from evidences)
•
Structure of an argument
First premises
Second premises
Third premises
Fourth premises
.
.
Nth premises
______________
∴ Conclusion
Valid Argument: It can be
described as an argument
where if all their premises
are true, then their
conclusions will also be true.
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Rules of Inference
Rules of Inference
(Deriving Conclusion from evidences)
Structure of an argument
• For example:
"If tomorrow is holiday, I will go to mall."
"tomorrow is holiday".
“therefore, I will go to mall."
P: tomorrow is holiday
Q: I will go to mall
•
P→Q
P
____________
∴Q
Designed By Team : Discrete Mathematics
Rules of Inference
Rules of Inference
(Deriving Conclusion from evidences)
Modus Ponens (Law of Detachment)
Suppose there are two premises, P and P → Q.
Now, we will derive Q with the help of Modules
Ponens like this:
P→Q
P
____________
∴Q
•
Example:
•
Suppose P → Q = "If we have a bank account, then we can take advantage of
this new policy.“
•
•
P = "We have a bank account."
Therefore, Q = "We canDesigned
take advantage
of this new policy."
By Team : Discrete Mathematics
Rules of Inference
•
Modus Tollens ( Law of Contraposition)
Suppose there are two premises, P → Q and ¬Q. Now, we will
derive ¬P with the help of Modules Tollens like this:
P→Q
•
¬Q
____________
∴ ¬P
• Example:
• Suppose P → Q = "If we have a bank account, then we can
take advantage of this new policy."
• ¬Q = "We cannot take advantage of this new policy."
Designed
By Team
: Discrete
• Therefore, ¬P = "We
don't
have
a Mathematics
bank account."
Rules of Inference
Rules of Inference
(Deriving Conclusion from evidences)
•
•
Hypothetical Syllogism
Suppose there are two premises, p → q and q → r. Now, we
will derive p → r with the help of Hypothetical Syllogism like
this:
p→q
q→r
____________
∴p→r
Example:
•
Suppose P → Q = "If my fiancé comes to meet me, I will not go to office."
•
Q → R = "If I will not go to office, I won't require to do office work."
•
Therefore, P → R = "If my fiancé come to meet me, I won't require to do
Designed By Team : Discrete Mathematics
office work."
Rules of Inference
Rules of Inference
(Deriving Conclusion from evidences)
•
•
Disjunction Syllogism (Rule of Presentation)
Suppose there are two premises ¬P and P ∨ Q. Now, we will
derive Q with the help of Disjunction Syllogism like this:
¬P
P∨Q
____________
∴Q
•
Example:
•
Suppose P ∨ Q = “I will make a tea or coffee."
¬P = “I will not make a tea."
Therefore, Q = “Therefore, I will make a coffee."
•
•
Designed By Team : Discrete Mathematics
Rules of Inference
Rules of Inference
(Deriving Conclusion from evidences)
•
•
Addition
Suppose there is a premise P. Now, we will derive P ∨ Q
with the help of Addition like this:
P
____________
∴P∨Q
Example:
• Suppose P be the proposition, "Harry is a hard working
employee" is true
• Here Q has the proposition, "Harry is a bad employee".
• Therefore, "Either Harry is a hard working employee Or
Harry is a bad employee".
Designed By Team : Discrete Mathematics
Rules of Inference
Rules of Inference
(Deriving Conclusion from evidences)
Simplification:
•Suppose there is a premise P ∧ Q. Now, we will
derive P with the help of Simplification like this:
P∧Q
__________
∴P
Example:
•Suppose P ∧ Q = "Harry is a hard working
employee, and he is the best employee in the
office".
•Therefore, "Harry is a hard working employee".
Designed By Team : Discrete Mathematics
Rules of Inference
Rules of Inference
(Deriving Conclusion from evidences)
Conjunction
•Suppose
there are two premises P and Q. Now, we will
derive P ∧ Q with the help of conjunction like this:
P
Q
___________
∴P∧Q
Example:
•Suppose P = "Harry is a hard working employee".
•Suppose Q = "Harry is the best employee in the office".
•Therefore,
"Harry is a hard working employee and Harry is
the best employee in the office".
Designed By Team : Discrete Mathematics
Rules of Inference
Rules of Inference
(Deriving Conclusion from evidences)
Resolution
•Suppose there are two premises P ∨ Q and ¬P ∨ R. Now, we will derive
Q ∨ R with the help of a resolution like this:
P∨Q
¬P∨R
____________
∴Q∨R
•Example:
•P ∨ Q = "my friend come to meet me, I will not go to office ".
•¬ P ∨ R = "my friend did not come to met me, I won't require to do
office work".
•Therefore, Q ∨ R = "I will not go to office or I won't require to do office
work".
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Designed By Team : Discrete Mathematics
Arya Studies hard, if Arya study hard
then he is dull.
If Arya is dull he will not get good job.
Hence Arya will not get the job. For solving such problems consider any two
premise first and add the conclusion of two
premise with the third premise.
Sol: Let, p: Arya studies hard.
q: He is dull.
r: He will get the job.
1. P
2. q
p→q
q→~r
p
--------------p→q
q (MP)
~r (MP)
q→~r
-----------Statement is valid
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~r
Designed By Team : Discrete Mathematics
•
Find weather the given statements are valid
•
Eg 1. if my health is good, I will study
My health is good.
Therefore, I will study
2. If I am 18+ then I can vote.
I can’t vote.
Therefore, I am 18+.
3. I will work or sleep.
I will not work
Therefore, I will sleep.
4. If I know English then I can write an essay.
If I write essay, I will get grades.
Hence, if I know English, I will get grades.
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For practice
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Determine whether the following argument is valid or not.
Eg: if eva wakes up, she can watch sunrise.
But eva is slept, she will therefore miss the sunrise.
Sol: p: eva wakes up.
q: she watch sunrise.
The rule of inference is
•
p→q
~P
-------------
hence its INVALID
∴ ~q
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Methods of Proof
Direct Method of Proof:
• Direct method of proof p->q is logical valid which start with
assumption that p is true & using this assumption we
consider q is true.
1. Get the direct proof that product of two odd integer is odd.
Sol: Let, x and y be the odd integers.
X=2n+1 & y=2m+1
Since, by putting any value in m and n we get x,y odd.
x.y= (2n+1)(2m+1) = 4mn+2n+2m+1
=2(2mn+n+m)+1 ∵ 2mn+n+m = a
=2.a + 1___odd integer.
Hence, the result follows.
•
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Methods of Proof
Direct Method of Proof:
• Direct method of proof p->q is logical valid which start
with assumption that p is true & using this assumption
we consider q is true.
2. Show that square of even number is even.
Sol: Let, x be an even number.
x=2n
Now, X2 = 4n2 = 2.(2n2)
•
=2. a
∵ n^2= a
->Hence, Result follows
Designed By Team : Discrete Mathematics
Methods of Proof
Direct Method of Proof:
• Direct method of proof p->q is logical valid which start
with assumption that p is true & using this assumption
we consider q is true.
3. Show that sum of two odd numbers is even.
Sol: Let, x & y be the two odd numbers.
y=2n+1, y= 2m+1
x+y= 2n+1 +2m+1 = 2n+2m+2
=2(m+n+1)
= 2.a
∵ (m+n+1)= a
•
=Even Number
=Hence, Result Follows.
Designed By Team : Discrete Mathematics
Methods of Proof
Indirect method: Method of contrapositive
• P->Q = ~Q -> ~P
1. Show that if n2 is odd then n is odd
Sol: Let, p:n2is odd, q: n is odd
~p : n2 is even, ~q: n is even
n=2x, x ∈ z
n2 = 4x2
=2.2.n2
= 2.a } hence, n2 is even
It means, ~Q->~P is true, which is logically
equivalent with p->q
•
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For In class Practice
Que 1. Prove that, If n is even, n2 is also even. Solve using
direct and contrapositive method.
Que 2. for all integers if both m and n are even, then m.n is
even. Solve using direct method.
Rewrite the following statements in symbolic form using
quantifiers
1. All students have taken SHD Course.
2. There is boy student in the class who is sports person.
3. Some students are intelligent but not hardworking.
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For In class Practice
•
Que 1. Prove that n3+2n is divisible by 3 using mathematical
induction.
•
Sol:
•
Step 1: Base: Consider given expression is true for n=1
•
13+2.1 =3 -------its divisible by 3.
•
Step 2: Induction: Put n=k in given expression & consider its true. (k3+2k)
•
Now put n=k+1 in given expression,
•
(k+1) 3 + 2(k+1) = k3 + 3k2+3k+1 +2k+2 = k3 + 3k2+5k +3
•
= k3 + 3k2+2k + 3k +3
•
= k3 +2k+ 3k2+ 3k +3 = Assume that, yellow highlighted part is
divisible by 3 and green part can be written as 3(k2+ k +1) so
anything multiple of 3 is also divisible by 3.
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For In class Practice
•
Que 2. Prove that (ab)n=anbn is true for every natural number
using mathematical induction
•
Sol:
•
Step 1: Base: Consider given expression is true for n=1
•
(ab)1=a1b1 --------- Hence step 1 is satisfied.
•
Step 2: Induction: Put n=k in given expression & consider its true.
•
(ab)K=akbk ……………(i)
•
Now put n=k+1 in given expression,
•
(ab)k+1= a(k+1)b(k+1)
•
(ab)k (ab) = a(k+1)b(k+1)
•
akbk (ab) = a(k+1)b(k+1)
•
a(k+1)b(k+1) = a(k+1)b(k+1)
•
=
Hence Step 2 is verified.
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For In class Practice (Mathematical Induction)
•
Que 3. Prove that 5n-1 is divisible by 4 for n>=1
•
Que 4. Prove that (n-1) 3+ n3+(n+1) 3 is divisible by 9.
•
Que 5. Prove that 8n- 3n is multiple of 5 for n>=1
Designed By Team : Discrete Mathematics
MIT School of Computing
Department of Computer Science & Engineering
PLD
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ByTeam
Team :: Discrete
Mathematics
Designed
Discrete
Mathematics
MIT School of Computing
Department of Computer Science & Engineering
PLD
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ByTeam
Team :: Discrete
Mathematics
Designed
Discrete
Mathematics
MIT School of Computing
Department of Computer Science & Engineering
PLD
DesignedBy
ByTeam
Team :: Discrete
Mathematics
Designed
Discrete
Mathematics
MIT School of Computing
Department of Computer Science & Engineering
PLD
DesignedBy
ByTeam
Team :: Discrete
Mathematics
Designed
Discrete
Mathematics
MIT School of Computing
Department of Computer Science & Engineering
PLD
DesignedBy
ByTeam
Team :: Discrete
Mathematics
Designed
Discrete
Mathematics
MIT School of Computing
Department of Computer Science & Engineering
PLD
DesignedBy
ByTeam
Team :: Discrete
Mathematics
Designed
Discrete
Mathematics