Question

Simplify each of the following expressions by applying one of the theorems.
(a) X′Y′Z + (X′Y′Z)′
(b) (AB′+ CD)(B′E + CD)
(c) ACF + AC′F
(d) A(C + D′B) + A′
(e) (A′B + C + D)(A′B + D)
(f) (A + BC)+(DE + F)(A + BC)′

Answer 1

Does this come under boolean algebra?

Answer 2

yes

Answer 3

please help me to answer this

Answer 4

What is
A+A'=?

Answer 5

1

Answer 6

In the first question
(X'Y'Z)' is the complement of (X'Y'Z)
do u see any relation between that theorem and ur first question?

Answer 7

how will i know if there is a relation between a theorem?

Answer 8

A+A'=1 right?
If so
what is
assume (X'Y'Z) as A
(X'Y'Z)+(X'Y'Z)'=?

Answer 9

1

Answer 10

yes:)

Answer 11

?

Answer 12

but i need to show a solution in my paper how i get that answer.

Answer 13

u can state the theorem u used to get the answer.
A+A'=1
It's complement law isnt it?

Answer 14

yes it is

Answer 15

so u can write
by complement law
(X'Y'Z)+(X'Y'Z)'=1

Answer 16

thanks. can you help me with the other questions?

Answer 17

ya.
For the second u need remove the brackets. for that u have to use distributive law.
(AB′+ CD)(B′E + CD)=AB'(B'E+CD)+CD(BE'+CD)

Answer 18

AB'(B'E+CD)+CD(BE'+CD)=?

Answer 19

X(Y+Z)=XY+XZ <<-- this ?

Answer 20

yes. using that expand
AB'(B'E+CD)+CD(BE'+CD

Answer 21

i dont know how. im soory

Answer 22

AB'(B'E+CD)+CD(BE'+CD=AB'*B'E+AB'*CD+CD*BE'+CD*CD

Answer 23

nw see if u can use this law in that
A*A=A

Answer 24

sorry if im to slow to reply. coz im trying to solve what law u give :)

Answer 25

that's k:)

Answer 26

are you busy? coz im having a trouble solving it. :(

Answer 27

nope i am free. take ur time and tell me where u r stuck>

Answer 28

i dont know w/c i will use in

Answer 29

the law that u give to me.

Answer 30

how i will i know what will be the given in A*A?

Answer 31

AB'(B'E+CD)+CD(BE'+CD=AB'*B'E+AB'*CD+CD*BE'+CD*CD
Did u get this?

Answer 32

in CD*CD u have CD instead of A that u had in A*A. following me?

Answer 33

how did u get BE'? IS THAT B'E?

Answer 34

oops sorry. yes it is B'E. am really sorry

Answer 35

no worries. :) lets proceed

Answer 36

ha k:)
in CD*CD u have CD instead of A that u had in A*A.
so if A*A=A
then
CD*CD=CD.
following me?

Answer 37

yah

Answer 38

AB'(B'E+CD)+CD(B'E+CD=AB'*B'E+AB'*CD+CD*B'E+CD*CD
=AB'B'E+AB'CD+CDB'E+CD
is this k?

Answer 39

ohhhh.. i get it :)

Answer 40

consider CDB'E+CD
can u factor out anything?

Answer 41

cd?

Answer 42

ya. factor it out and simplify

Answer 43

CDB'E+CD in this given i will factor out and simplify?

Answer 44

ya.
CDB'E+CD
factor out
CD(B'E+1)
A+1=1
so
CD(B'E+1)=CD*1=CD.getting it?

Answer 45

yah

Answer 46

AB'B'E+AB'CD+CDB'E+CD=AB'B'E+AB'CD+CD
=AB'E+CD

Answer 47

how will i know if i will use the laws?

Answer 48

AB'CD+CD
check again. if u can factor out anything. yes. u can factor CD
Then when u factor out CD u will get
CD(AB'+1)
u knw A+1=1
u have AB' instead of A.
So
AB'+1=1
Then
CD(AB'+1)=CD
AB'B'E
as u can see
B'B' is similar to AA=A
so
B'B'=B
so
AB'B'E=AB'E.
so finally
AB'B'E+AB'CD+CD=AB'E+CD

Answer 49

ok. gets

Answer 50

hw abt trying this nw?
ACF + AC′F

Answer 51

try only . but not sure.

Answer 52

tell me ur answer

Answer 53

can u start first then i will follow?

Answer 54

try factor out the common thing in it

Answer 55

af(?????)

Answer 56

yup

Answer 57

uhmm. im having a trouble with the c' and c. i dont know what will i write

Answer 58

ACF+AC'F=AF(C+C')

Answer 59

(C+C')
does this look similar to any laws u knw?

Answer 60

1?

Answer 61

yup:)
so
what is
AF(C+C')

Answer 62

af

Answer 63

yup good:)

Answer 64

hehe.

Answer 65

(A′B + C + D)(A′B + D)=A'B(A'B+D)+C(A'B+D)+D(A'B+D)
is this right??

Answer 66

yup:)

Answer 67

expand it and simplify:)

Answer 68

=a'b*a'b+a'b*d+c*a'b+c*d+d*a'b+d*d
=a'ba'b+a'bd+c'ab+cd+da'b+d
=a'ba'b+a'bd+ca'b+cd+d
am i right?

Answer 69

yup:) u can simplify it further

Answer 70

uhm. thats my problem now. hehe.

Answer 71

a'ba'b=?

Answer 72

a'b

Answer 73

yup:)

Answer 74

cd+d=?

Answer 75

d?

Answer 76

yup:)

Answer 77

a'ba'b+a'bd+ca'b+cd+d=?

Answer 78

a'b+d
???hehe

Answer 79

yup:)

Answer 80

but what happen to the a'bd+ca'b?

Answer 81

a'ba'b+a'bd+ca'b+cd+d=a'b+a'bd+ca'b+d
=a'b(1+d)+ca'b+d
=a'b+ca'b+d
=a'b(1+c)+d
=a'b+d

Answer 82

can you still help me with the last 2 questions?

Answer 83

let me try them out and tell u

Answer 84

ok. thanks

Answer 85

(e) (A′B + C + D)(A′B + D)

( A′B + D + C)(A′B + D)
------- --------

say A'B + D = X

(X + C)X

XX + CX

X + CX

X(1+C)
X

A'B + D

Answer 86

(f) (A + BC)+(DE + F)(A + BC)'
------- --------

say A+BC = X, DE + F = Y

X + X'Y

(X+X')(X+Y)

1(X+Y)

X + Y

A + BC + DE + F

Answer 87

A(C+D'B)+A'=AC+AD'B+A'
=AC+AD'B+A'.1 (C+1=1)
=AC+AD'B+A'(C+1)
=AC+AD'B+A'C+A'
=C(A+A')+AD'B+A'.1
=C+AD'B+A'(DB'+1) (D'B+1=1)
=C+AD'B+A'D'B+A'
=C+D'B(A+A')+A'
=C+D'B+A'

Answer 88

another way for D :
(d) A(C + D′B) + A′

say X = A', C+D'B = Y

X'Y + X

(X'+X)(X+Y)

X + Y

A' + C + D'B

Answer 89

hi rational im a bit having a trouble with the last part of the solution u had. but it is good coz it is short way to solve.

Answer 90

below distributive law is a powerful property :
\(x+yz = (x+y)(x+z)\)

Answer 91

hi ajprincess can u give me another way to solve letter f?

Answer 92

e and f uses that property, so if you know how to use that property there should not be any problems

Answer 93

I totally forgot about that law. thanx a lot @rational. i remembered it nw cos of u.:) well alternative way to solve f would be the one i used for d part. bt I guess it will be too lengthy.

Answer 94

thanks @ajprincess i learned a lot. hope that u can help me again. @rational thanks for solving alternative way to solve :)

Answer 95

U r welcome:) glad to knw that u learnt a lot :)

Answer 96

hi @ajprincess may i ask a question. in this part >>> =AC+AD'B+A'.1 (how you get A'.1?)

Answer 97

when we multiply a number or variable by 1 the answer is the same as number or variable.
i.e
A*1=A
3*1=3
I used the converse part