de morgans
TRANSCRIPT
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Propositional logicThe propositional logic representlogic through proposition and
logical connectives.
We may define proposition as an
elementary atomic sentence that
may take either true value orfalse value but may not take
other value .
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Consider the following examples :
It is raining [it is a proposition as it
may either be true orfalse .
Australia won the ICCworld cup 2007.
It is a also proposition asit is true.
India is a continent It is a proposition as it isfalse
What did you eat It is not proposition as it
does not result in true orfalse.
How are you Not a proposition for the
similar reason as above
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A proposition is an
elementary atomic
sentence that may either
be true or false but may
not take other value .
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Proposition are also
called sentence orstatement after this
introduction , let us talkabout terms and symbol
used in propositionallogic .
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A symbol proposition is one thatdoes not contain any other
proposition of part . We will usethe lower case letter p, q, r., as
symbols for simple statements or
proposition .
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A compound proposition is one
with two or more simpleproposition as parts or what
We will call components . Acomponent of compound is any
whole proposition that is a parts
of larger proposition ; componentsmay themselves be compounds.
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For example , following arecompound proposition :
it is raining and wind is blowing .Take it or leave it .
If you work hard then you will
rewarded.
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An operators ( or connective ) joins simpleproposition into compounds, and joins
Compound into larger compounds. We will usethe symbols, +, ., and todesignate the sentential connectives . They arecalled sentential connectives because they join
sentences ( or what we are calling statement orpropositions ) . The symbol , ~ , is the onlyoperator that is not a connective ; it affectssingle statement only , and does not joinstatements into compounds.
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Different types of
connectives (oroperators ) used in
propositional logic are
given below :
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Represent by symbols + or v.
Disjunction means one of the
two arguments is true or bothe.g., p + q (or p v q) means p OR
q. It`s meaning is either p istrue , or q is true , or both
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Represent by symbols . or &
or . conjunction means both
arguments are true e.g., p . q(or p & q) means p and q. Its
meaning is both p and q aretrue .
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Represented x by symbols or
or
Implication means if one argument istrue then other argument is true e.g. , p
q(or p q or p q) means if p
then q . Its meaning is if p is true , then
q is true .
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Represented by symbols or .
Equivalence (or bi - conditional ) means
either both argument true or both false ,e.g., p q (or p q ) means if and
only if p is true then q is true . Its
meaning is p q are either both true or
both false .
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Not a connective actually , just anoperator . Represented by - or` or
~ (bar).
It is an operator that effects asingle statement only and does not
join two or more statement e.g., ~p (or p` or p ) means NOT P . Its
meaning is P Is False .
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A truth table is a
complete list ofpossible truth
values of a
proposition
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The not operators
works on singleproposition , thus it is
also called unaryconnective sometimes .
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If p denotes proposition , then
its negation will be denoted by~ p or q or p . If p is 0 (false),then ~ p is 1 (true ) and if p is 1
(true) then ~ p is 0 (false) . Thetruth table for this operation is
shown as follows:
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P P
0 1
1 0
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Also note that
NOT (NOT p) result into p itself
i.e.,
p = p
or (p) = por ~ (~p) = p.
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The OR connective workswith more then one
proposition . The compound p+ q has two (2) components
proposition (p and q), each ofwhich can be true or false .
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So there are (2* 2) possible
combination . The disjunctionof p with q (denoted as p + q
or p v q ) will be truewhenever p is true or q is
true or both are true .Consider the truth table given
below :
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p q p + q
0 0 0
0 1 1
1 0 1
1 1 1
TRUTH TABLE FOR DISJUNCTION (OR)
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The AND connective also
works with more than one
proposition compound The p q (or p & q) will be true
whenever both p and q aretrue
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p q p - q
0 0 0
0 1 0
1 0 0
1 1 1
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In the conditional p qthe first proposition (the if -
clause ) p here , is called the
antecedent and the second
proposition (then clause ) q
here , is called consequent.
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In the more complex conditionals
, the antecedent and theconsequent could themselves be
compound proposition . Theconditional p q will be false
when p is true and q is false . For
all other input combinations, itwill be true
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p q p q
0 0 1
0 1 1
1 0 0
1 1 1
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The conditional p q
may be expressed as follow:P q = p` + q
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A bi conditional results into false
when one of its component
proposition is true and other is false.
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That is, p q will
be 0 (false ) when p is 0and q is 1 or p is 1 and q
is 0 . For all other inputs ,p q is 1.
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p q p q
0 0 1
0 1 0
1 0 0
1 1 1
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The conditional p q may be
expressed as :p q = pq + p` . q`
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The proposition that
have some combinationof 1`s and 0`s in their
truth table columns,are called cotingencies
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The propositionshaving nothing but
1`s in their truth
table column, are
called tautologies.
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The propositions having
nothing but 0 `s in arecalled their truth table
column, are calledcontradictions
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Two statements are
consistent if and onlyif their conjunction is
not a contradiction .
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The converse of a conditional
proposition is determined by
interchanging the antecedentand consequent of given
conditional ..
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It results into new
conditional . e.g .,
converse of p q
is q p
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That is, if
p : it is raining.q : sky is not clear
.then , p q = if it
is raining then sky is not
clear .
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it`s converse will be new
conditional as givenbelow :
q p = if sky is not
clear then it is raining .
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The inverse of a conditional proposition
is another conditional having negated
antecedent and consequent . That is ,the inverse of p q is p` q` .
E.g., if
p : it is raining.q : sky is not clear .
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then , p q = if it is
raining then sky is not clear
it`s inverse will be a new
conditional as given below :
p` q` = if itis not raining then sky is not
clear .
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The cotnrapositive of a conditional isformed by creating another conditionalthat takes its antecedent as negated
consequent of earlier conditional andconsequent as antecedent of earlierconditional . That is , contrapositive
p q is ~q ~p or q por q` p`.
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1. 0 + P= P
0 . P = 0
Properties of 0
2. 1 + p = 11 . p = p
Properties of 1
3 . p + pq = p
p + (p + q)= p
absorption law .
4. p = p involution
5. p + p =pp . p = p
Idempotence law
6. p + p = 1p . p = 0
Complementaritylaw
7. p + q = q + pp . q = q . p
commulative law
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8. (p +q ) + r = p + (q + q)
(p . q ) . r = p . (q. r)
Associative Law
9.p . (q + r ) = (p .q) + (p .r)
p + (q . r ) = (p +q ). (p + r)p + p q = p + q
Distributive Law
10 . p + q = p . qp . q = p + q
De Morgan`s Law
11. p q = p + q Conditional
Elimination
12. p q = (p q ) . (q p ) Bi - ConditionalElimination
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Construction a truth table for the expression(A . ( A + B) ). What single term
Is the expression equivalent to ?
A B A + B (A . (A + B))
0 1 0 0
0 1 1 0
1 0 1 1
1 1 1 1
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Using truth table, prove p qis equivalent to ~ p ~ q
p q ~q ~ p p q ~ p ~q
0 0 1 1 1 1
0 1 0 1 1 1
1 0 1 0 0 0
1 1 0 0 1 1
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p q p p q p + q
0 0 1 1 1
O 1 1 1 1
1 0 0 0 0
1 1 0 1 1
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Prove that p q = q p
p q p q q p
0 0 1 1
0 1 0 0
1 0 0 0
1 1 1 1
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Prove that p q = (q p) . ( q p ).
P q p q p q q p (p q) . (q p)
0 0 1 1 1 1
0 1 0 1 0 0
1 0 0 0 1 0
1 1 1 1 1 1
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Consider some simple proposition
A : it is raining.B : wind is blowing .
C : I am not driving.
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1. A V B : it is raining and wind isblowing
2. ~ B : wind is not blowing .3. ~ A . C : It is not raining and I
am not driving ,
4. A. ~ c : it is raining and I amnot driving
5. A + B . C : it is raining or wind
is blowing and I amnot driving .
Prove that x + 1 is a tautology
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Prove that x + 1 is a tautology.
x 1 x + 1
0 1 1
1 1 1
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Prove that x + x ` is a tautology andx . X ` is a contradiction
x x` x + x` x . X`
0 1 1 0
1 0 1 o
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Before we start discussion abo utlogical operators, let us first
understand what a truth table is.
For example, following logicalstatements can have only one of the
two values (TRUE (YES) orFALSE (NO))
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1.I WANT TO HAVE
TEA.2.TEA IS READLY
AVABILABLE.
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Truth table is a table which
represents all the possiblevalue of logical variables /
statements along with all thepossible results of the given
combinations of values
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X Y R
1 1 1
1 0 0
0 1 0
0 0 0
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1Represent true value and 0
represent false value .
2This is a truth table i,e.,table of truth values of truth
functions.
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If result of any logical
statement or expression is
always TRUE or I , it iscalled Tautology and if the
result is always FALSE or 0it is called
FALLACY
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Now let us proceed with our
discussion about logical operators I,e.,
1. NOT operator2. OR operator3. AND operator
O O O
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NOT OPERATORthis operator operates onsingle variable and
operation performed by notoperator is called
complementation and thesymbol we use for it (bar) .
Thus x means complement of x
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Thus x means complement of x
and yz means complement of y z
. As we know, the variables used inBoolean equations have unique
characteristics that they may
assume only one of two possiblevalues 0 and 1 , where 0 denotes
FALSE and 1 denotes TRUE value. Thus the complement operation
can be defined quit simply .
0 = 1
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0 = 1
1 = 0
Truth table for not operator
x x
0 11 0
S l h b l l
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Several other symbol e.g ~ , are also usefor complementation symbol . If ~ is used
then ~ x is read is negation of x and ifsymbol is used then x is read complementof x .
x
x x
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Not operation is singular
or unary operation as itoperates on single variable .
Venn diagram for x is givenabove where shaded area
depicts x
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A second important
operator in Boolean algebrais or operator which denotes
operation called logical
addition and the symbol weuse for it is + .
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0 + 0 = 0
0 + 1 = 11 + 0 = 1
1 + 1 = 1
T th t bl f OR t
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Truth table for OR operator
X Y X + y0 0 0
0 1 11 0 1
1 1 1
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AND operator perform another
important operation of Boolean
algebra called logical multiplicationand the symbol for AND operation
is (.) dot . Thus x . Y will be
read as x AND Y . The rules forand operation are :
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0. 0 = 0
0 . 1 = 01. 0 = 0
1 . 1 = 0
T th t bl f AND t
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Truth table for AND operator
X Y X . Y
0 0 00 1 0
1 0 01 1 1
V di f X Y i i b l
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Venn diagram for X . Y is given belowwhere shaded area depicts (X .Y)
YX
Shaded portionshows
X . Y
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Logical variables are
combined by mean of logical
operation (AND, OR, NOT )
to form Boolean expression .
For example , x + y . Z + zis a Boolean expression .
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X Y Z0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
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X Y Z Y . Z
0 0 0 0
0 0 1 0
0 1 0 0
0 1 1 1
1 0 0 0
1 0 1 0
1 1 0 0
1 1 1 1
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X Y Z Y . Z Y Z0 0 0 0 1
0 0 1 0 1
0 1 0 0 10 1 1 1 0
1 0 0 0 1
1 0 1 0 11 1 0 0 1
1 1 1 1 0
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X Y Z Y . Z YZ X + YZ
0 0 0 0 1 1
0 0 1 0 1 1
0 1 0 0 1 1
0 1 1 1 0 0
1 0 0 0 1 1
1 0 1 0 1 1
1 1 0 0 1 1
1 1 1 1 0 1
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X Y XY X + XY
0 0 0 0
0 1 0 0
1 0 0 11 1 1 1
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X Y X+Y (X+Y) X Y XY
0 0 0 1 1 1 10 1 1 0 1 0 0
1 0 1 0 0 1 01 1 1 0 0 0 0
PREPARE A TABLE OF COMBINATIONS FOR THE FOLLOWING BOOLEAN
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PREPARE A TABLE OF COMBINATIONS FOR THE FOLLOWING BOOLEANALGEBRA EXPRESSIONS.
1.XY+XY
2.XYZ+XYZ
3.XYZ+XY
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1. x y + x y is a 2 variable expression , its truth table asfollows :
X Y X Y X Y X Y X Y+ XY
0 0 1 1 1 0 1
0 1 1 0 0 1 1
1 0 0 1 0 0 0
1 1 0 0 0 0 0
b XYZ + X Y Z
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b. XYZ + X Y Z
Truth Table
X Y Z X Y Z XYZ X Y Z XYZ + X Y Z
0 0 0 1 1 1 0 0 0
0 0 1 1 1 0 0 1 1
0 1 0 1 0 1 0 0 0
0 1 1 1 0 0 0 0 0
1 0 0 0 1 1 0 0 0
1 0 1 0 1 0 0 0 0
1 1 0 0 0 1 1 0 1
1 1 1 0 0 0 0 0 0
c. X Y Z + X Y
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Truth TableX Y Z X Y Z X Y Z XY X Y Z + XY
0 0 0 1 1 1 0 0 0
0 0 1 1 1 0 0 0 0
0 1 0 1 0 1 1 0 1
0 1 1 1 0 0 0 0 0
1 0 0 0 1 1 0 1 11 0 1 0 1 0 0 1 1
1 1 0 0 0 1 0 0 0
1 1 1 0 0 0 0 0 0
A. X (Y + Z ) + X Y
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Truth Table
X Y Z Y Z ( Y +Z) X (Y + Z) XY X (Y + Z )+ XY0 0 0 1 1 1 0 0 0
0 0 1 1 0 1 0 0 0
0 1 0 0 1 1 0 0 0
0 1 1 0 0 0 0 0 0
1 0 0 1 1 1 1 1 1
1 0 1 1 0 1 1 1 1
1 1 0 0 1 1 1 0 1
1 1 1 0 0 0 0 0 0
B. XY (Z + YZ ) + Z Truth Table
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B. XY (Z YZ ) Z Truth Table
X Y Z Y Z YZ Z + YZ XY XY(Z +YZ) XY (Z + YZ )+ Z
0 0 0 1 1 0 0 0 0 1
0 0 1 1 0 0 1 0 0 0
0 1 0 0 1 1 1 0 0 1
0 1 1 0 0 0 1 0 0 0
1 0 0 1 1 0 0 1 0 1
1 0 1 1 0 0 1 1 1 1
1 1 0 0 1 1 1 0 0 1
1 1 1 0 0 0 1 0 0 0
C . A [ (B + C ) + C ] Truth Table
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C . A [ (B + C ) + C ] Truth Table
A B C B C ( B + C ) ( B + C) + C A [ (B + C) + C ]
0 0 0 1 1 1 1 0
0 0 1 1 0 1 1 0
0 1 0 0 1 0 1 0
0 1 1 0 0 1 1 0
1 0 0 1 1 1 1 1
1 0 1 1 0 1 1 1
1 1 0 0 1 0 1 1
1 1 1 0 0 1 1 1
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Gates are digital (two - state )circuits because the input and
output signals are either low
voltage ( denotes 0 ) or highvoltage (denotes 1 ). Gates are
often called logic circuits becausethey can be analyzed with Boolean
algebra
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A Gate is simply an
electronic circuit whichoperates on one or
more signals t o producean output signal .
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Inverter (NOT gate )
OR gateAND gate
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An inverter (NOT Gate ) is
a gate with only one inputsignal and one output
signal ; the output state isalways the opposite of the
input state
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An inverter is also called a NOT
gate because the output is not
the same as the input . The
output is sometimes called thecomplement (opposite) of the
input . Following tablessummaries the operation:
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X YLow HighHigh Low
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X Y
0 1
1 0
A low input I e 0 produce high output
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A low input I,e., 0 produce high output
I,e. 1, and vice versa. The symbol for
inverter is given in:
x x
NOT GATE SYMBOL
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The OR Gate has two or moreinput signals but only one
output signal . If any of theInput signals is 1 (high ). The
output signal is 1 (high)
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If all inputs are 0
then output is also 0. If one or more
inputs are 1, the
output is 1 TWO INPUT OR GATE
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X Y F
0 0 00 1 1
1 0 11 1 1
F = X + Y
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X Y Z F
0 0 0 0
0 0 1 10 1 0 1
0 1 1 1
1 0 0 1
1 0 1 1
1 1 0 1
1 1 1 1
F = X + Y + Z
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A
B
F
Two input OR gate
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AB FC
Three input OR gate
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A
B FC
Four input OR gate
D
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The AND Gate can have two ormore then two input signals
and produce an output signal .When all the input are 1 i.e,
that is high then output is 1
otherwise output is 0 only .
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X Y F0 0 0
0 1 01 0 0
1 1 1
F X Y
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X Y Z F
0 0 0 0
0 0 1 0
0 1 0 0
0 1 1 0
1 0 0 0
1 0 1 01 1 0 0
1 1 1 1
F = X Y Z
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2 Input AND gate
A
BF
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3 Input AND gate
A
B F
C
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4 Input AND gate
A
B FCD
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Boolean algebra being a system of
mathematics consists of fundamentallaws that are used to build a
workable cohesive framework uponwhich are based the theorems of
Boolean algebra . These fundamental
laws are known as Basic postulatesof Boolean
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1. if x = 0 then x = 1; and if x = 1 then x = 0
2. OR Relation s (Logical Addition)
0 + 0 = 0 00
0 + 1 = 1
1 + 0 = 1
1 + 1 = 1
OR
OR
OR
OR
0
1
1
1
01
1
0
11
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0 . 0 = 0
0 . 1 = 1
1 . 0 = 1
1 . 1 = 1
AND
AND
AND
AND
0
0
0
1
0
1
11
0
0
1
0
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0 = 1
1 = 0
0
01
1
NOT
NOT
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This is a very important principle used inBoolean algebra . This state s that startingwith Boolean relation can be derived by .
1. changing each OR sign ( + ) to an AND sign(.)
2. changing each AND sign ( . ) to an OR sign
(+)in3. Replacing each 0 by 1 and each 1 by 0 .
The derived relation
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using dualityprincipal is called
dual Of originalexpression.
For instance , we take postulate II related
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to logical addition , which states
A. 0 + 0 = 0B. 0 + 1 = 1
C. 1 + 0 = 1
D. 1 + 1 = 1
Now working according to above
d l h d d
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guidelines , + is changed to. And 0
s are replaced by 1s , these becomeA. 1 . 1 = 1
B. 1 . 0 = 0C. 0 . 1 = 0
D. 0 . 0 = 0
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1. 0 + x = x
2. 1 + x = 1
3. 0 . x = 0
1 x x
OR 00
0
OR 11
X
AND 1
0
X
AND X1
X
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0 X R
0 0 0
0 1 1
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1 X R
1 0 1
1 1 1
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0 X R
0 0 0
0 1 0
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1 X R
1 0 0
1 1 1
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OR
X
X
X
A. X + X = X i,e;
(gate representation for (a))
B. X . X = X i,e;
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,e;
AND XX
X
(gate representation for (b))
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X X R
0 0 01 1 1
0 + 0 = 01 + 1 = 1
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X X R
0 0 01 1 1
0 . 0 = 01 . 1 = 1
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(X) = X
X
X X = X
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X X R
0 1 0
1 0 1
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A. X + X = 1
OR
XX X + X = 1
( gate representation of (a)
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B. X . X = 0
ANDX
X X . X = 0
( gate representation of (b)
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X X X + X
0 1 1
1 0 1
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X X X . X
0 1 0
1 0 0
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A. X + Y = Y + X
ORX
Y
R = ORYXR
B. X . Y = Y . X
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ANDX
Y
R ANDYX
R
=
Truth Table X + Y = Y + X
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X Y X + Y Y + X
0 0 0 0
0 1 1 11 0 1 1
1 1 1 1
Truth Table for X . Y = Y . X
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X Y X . Y Y . X
0 0 0 0
0 1 0 01 0 0 0
1 1 0 1
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A. X + (Y + Z) = (X + Y ) + Z
ORYZ
X
OR =
R
Y + Z
ORXZ
Z
ORR
X + Y
B. X (YZ) = (X Y) Z
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( ) ( )
AND
Y
Z
X
AND =R
Y Z
ANDXY
Z
ANDR
X Y
Truth Table for X + (Y + Z) = (X + Y ) + Z
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X Y Z Y + Z X + Y X + (Y + Z ) (X + Y )+ Z
0 0 0 0 0 0 0
0 0 1 1 0 1 1
0 1 0 1 1 1 1
0 1 1 1 1 1 1
1 0 0 0 1 1 1
1 0 1 1 1 1 11 1 0 1 1 1 1
1 1 1 1 1 1 1
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A. X (Y+Z) = X Y+ XZ
OR
Y
Z
Y + Z
ANDX R
=
AND
AND
OR
X
Y
XY
XY
R
Z
B X +YZ = ( X +Y)( X+Z)
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B. X +YZ ( X +Y)( X+Z)
ANDY
Z
Y . Z
ORXR
=
OR
OR
AND
X
Y
X + Y
X+Y
R
Z
Truth Table for X + (Y + Z) = X Y + X Z
X Y Z Y Z X Y XZ X(Y Z ) XY XZ
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X Y Z Y + Z X Y XZ X(Y + Z ) XY + XZ
0 0 0 0 0 0 0 0
0 0 1 1 0 0 0 0
0 1 0 1 1 0 0 0
0 1 1 1 0 0 0 0
1 0 0 0 0 0 0 0
1 0 1 1 0 1 1 1
1 1 0 1 1 0 1 1
1 1 1 1 1 1 1 1
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A. X + XY = X
ANDY
Z
ORX Y
X
X
B X (X +Y) X
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B. X (X + Y) = X
ORY
ANDX+ Y
X
X
Truth Table for X + XY = X
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X Y XY X + XY
0 0 0 0
0 1 0 0
1 0 0 1
1 1 1 1
Column X and X + XY are identical . Henced Al i b d l b i ll
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proved . Also it can be proved algebraically
as
L. H.S = X + XY
= X (1 + Y)
Putting 1 + y = 1X . 1 = X = R.H.S
(b)Since rule (b) is dual of rule (a) it is also proved
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.
L.H.S. =X (X +Y) = X . X + XY
= X . X + XY
= X + XY= X ( 1 + Y )
= X . 1= X
= R. H.S
(X.X=X: Indempotence Law)
(using 1 + y =1 properties of 0 , 1)
(X .1 = X USING PROPERTIES OF0,1)
X XY X Y (Thi i h hi d di ib i l )
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X + XY = X + Y (This is the third distributive law )
AND
OR = OR
x
y
RRXY
y
X
NOT
X + XY = X + Y
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PROOF L.H.S = X + X Y= X . 1 + X Y= X (1 + Y ) + XY= X + XY + XY
= X + Y (X + X )= X + Y . 1= X + Y
R. H . S PROVED
(X. 1 = X PROPERTY OF 0 AND 1)
(1+Y= 1 PROPERTY OF 0 AND 1
(X +X =1 COMPLEMENTARITY LAW)
(Y .1 =Y property of 0 and 1)
1 0 + X = XPROPERTIES OF 0
2 0 . X = 0
3 1 + X = 1 PROPERTIES OF 1
4 1 X = X
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BOOLEAN ALGEBRA
RULES
4 1 . X = X
5 X + X = X INDEMPOTENCE LAW
6 X . X = X
7 X = X INVOLUTION
8 X + X = 1 COMPLEMENTARITY LAW
9 X . X = 0
10 X + Y = Y + X COMMULATIVE LAW
11 X . Y = Y.X
12 X + (Y + Z ) = (X + Y ) + Z ASSOCIATIVE LAW
13 X (YZ ) = (XY) Z
14 X (Y + Z ) = XY + XZ DISTRIBUTIVE LAW
15 X (Y + Z ) = (X + Y ) ( X + Z)
16 X + XY = X ABSORPTION LAW
17 X * ( X + Y ) = X
18 X + XY = X + Y
I h
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It states that x + y = x y
OR
ANDxy
R
NOT
X
Y
NOT
NOT
X
Y
R
PROOFTo prove this theorem , we need
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to recall complementarity laws, which
state thatX + X = 1 and X . X = 0
I,e, a logical variable / expression
when added with its complementproduces the output as 1 and when
multiplied with its complement
produces the output as 0 .
Now to prove DeMorgan s first theorem , we willuse complementarity laws .
let us assume that P = X + Y where , p , x , y
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are logical variables .
Then according to complementation law.P + P = 1 and P . P = 0
That means, if P , X , Y are boolean variablesthen this complementarity law must hold forvariables P . In other words , if p i.e X + Y = X Ythen
(x + y ) + x y must be equal to 1 .( x + y ) . X y must be equal to 0
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let us prove the first part , i, e.
(X + Y) + ( X Y ) = 1
( X + Y ) + X Y= ((X +Y ) + X) . ((X + Y ) + Y)= (X + X ) + Y) . (X + Y + Y)
= (1 + Y ) .( X + 1 )= 1 . 1= 1
(Ref. X + YZ = (X +Y ) (X + Z ))
(Ref . X + X = 1)(ref . 1 + x = 1)
So part is proved .Now let us proved second part I,e.
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( X + Y ) . X Y = 0( X + Y ) . X Y = X Y . ( X + Y)
= X Y X + X Y Y= X X Y + X Y Y= 0 . Y + X . 0= 0 + 0 = 0
So, second part is also proved , thus : X + Y = X Y
(ref X (YZ ) Z= (XY) Z)
(Ref . X (Y + Z )=XY + XZ )
(Ref . X .X = 0)
It t t th t
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It states that x . y = x + y
AND
ANDxy
R
NOT
XNOT
X
Y
RX . Y
NOT
Proof
If XY s complemrnt is X + Y then it must be true that
1. XY + ( X + Y ) = 1 AND
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2. XY ( X + Y ) = 0
L.H.S = XY + ( X + Y )= (X + Y + X ) . (X + Y + Y)= ( X + X + Y ) . ( X + Y + Y )
= ( 1 + Y ) . (X + 1 )= 1 . 1= 1 = R.H.S
(ref X + Y = Y + X )Ref.(X + Y )( X + Z ) = X + YZ)
ref X + X = 1ref 1 + x = 1
Now the second part i ,e
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L.H.S = XY . ( X + Y )= X Y X . X Y Y= X X Y . X Y Y
= 0 . Y + X . 0= 0 + 0 = 0 = R.H.SXY . ( X + Y) = 0 and XY + (X + Y ) = 1
ref X (Y + Z ) = XY + XZ
ref ( X . X = 0 )
XY . ( x + y ) = 0
Although the identifies above represent DeMorgan s theoremthe transformations more easily performed by following these step:
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y p y g p
1. complement the entire function2. change all the ANDs (.) to ORs (+) and all
the ORs (+) to ANDs (.)
3. complements each of the individual variables .
Break the line , charge the sign to demorganize aBoolean expression
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AB + A + AB = A + B + A + AB
= (A + B ) + ( A + AB )= ( A + B ) + (A + AB)
= A . B . (A . AB)= A . B ( A . ( A + B ))= AB (AA + AB)
= AB (0 + AB )= AB . 0 + ABAB = 0 + AABB= 0 + A A . 0= 0 + 0 = 0
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FOR EXAMPLE,X + X . Y = X
Its dual will be X . (X + Y ) = X
(Remember change .to + and viceversa; complement 0 and 1.)Similarly (X + Y) + Z = X + ( Y +Z )
is dual of (X . Y ) .Z = X . (Y . Z)X + 0 = X is dual of X . 1 = X
All boolean algebra is predicated on this two for one basis
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All boolean algebra is predicated on this two for one basis
Example : give dual of the following result in boolleansalgebra :
X . X = 0 for each X
Solution : using duality principal , dual of X . X = 0 isX + X = 1 (By changing (.) to ( +) and viseversa and by replacing 1s by 0s and viceversa)
Example : give the dual of X + 0 = X for each X .Solution : using duality principal , dual of X + 0 = X is
X .1 = X
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Example : state the principal of dulity in boolean algebra and give that dual of
the boolean expression :
(X + Y ). (X + Z ) . (Y + Z )Solution : principal of duality states that from every boolean relation, anotherboolean relation can be derived by
1. changing each OR sign ( +) to an AND ( .) sign2. changing each AND ( .) sign to an OR ( + ) sign3. replacing each 1 by o and each 0 by 1
The new derived relation is known as the dual of the original relation.Dual of ( X + Y) . ( X + Y ) . (Y + Z ) will be
(X . Y ) + ( X . Z ) + ( Y . Z ) = XY + X Z + Y Z
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