Showing posts with label Mathematics. Show all posts
Showing posts with label Mathematics. Show all posts

Nov 2, 2019

12 LAWS AND RULE OF BOOLEAN ALGEBRA

12 rules and laws boolean algera
basic rules of boolean algebra

BOOLEAN ALGEBRA'S RULES AND LAWS:

  A English mathematician, philosopher and logician George Boole 1854 given a mathematical theories and few logical algebra named as Boolean Algebra..which consist of 12 laws or rule which briefly described Boolean Algebra which are:
            
1.    A + 0 = A
2.    A + 1 = 1
3.    A . 0  = 0
4.    A . 1  = A
5.    A + A = A
6.    A + Ā = 1
7.    A . A = A
8.    A . Ā= 0
9.    ﬢ(ﬢA) = A
10. A + A B = A
11. A + ĀB = A + B
12. (A + B)(A + C) = A + BC

Derivation of Laws Using Gates And Truth Table:


RULE # 1 (  A + 0 = A) 

boolean simplification rules
A + 0 = A

A
0
  A+0
0
0
0
1
0
1

RULE # 2 (A + 1 = 1)

basic rules of boolean algebra
A + 1 = 1







A
1
  A+1
0
1
1
1
1
1

RULE # 3 (  A . 0  = 0)

boolean algebra rules pdf
A .0 =0






A
0
  A . 0
0
0
0
1
0
0

RULE# 4 (A . 1  = A)

logic algebra rules
A. 1= A







A
1
  A . 0
0
1
A
1
1
A

RULE #5 ( A + A = A)

12 rules of boolean algebra
A + A = A.







A
A
  A +A
0
0
A
1
1
A


RULE # 6(A + Ā = 1)


logic gates simplification rules
A + Ā = 1








A

Ā

 A+ Ā=1
0
1
1
1
0
1

RULE # 7(A . A = A)

boolean algebra laws and rules pdf
A.A=A






A
A
  A .A
0
0
A
1
1
A

RULE # 8 (  A . Ā= 0)

logic gate simplification rules
A . Ā= 0






A

Ā

 A . Ā
0
1
0
1
0
0

RULE # 9 (ﬢ(ﬢA) = A)

boolean expression simplification rules
A Double not = A







RULE #10 ( A + A B = A)

= A+ AB
= A (1 + B)
= A (1)             (from rule no 2 A + 1= 1, replace the variable A by B so  B+1 or 1+B=1)
= A
boolean algebra laws and rules
A + AB = A








RULE # 11( A + ĀB = A + B)

= A + ĀB
= A + AB + ĀB   (from rule no 10,A+AB=A)
= A + B(A +Ā )
= A + B(1)          (from rule no 6, A +Ā = 1)
= A + B              (which is equal to right hand side)

algebra laws and rules
 A + ĀB = A + B








RULE#12(. (A + B)(A + C) = A + BC)

= (A + B)(A + C)
= AA + AC + AB + BC
= A + AC + AB + BC        (from rule no 7 A.A = A)
= A(1 + C) + AB + BC
= A (1) + AB +BC             (from rule no 2, 1+C=1)
= A + AB + BC
= A + BC                           (from rule no 10 , A + AB = A) hence it is equal to right hand side
boolean equation rules
(A+B)(A+C)=A+AB











                                           physics label                                                                                                  

Oct 1, 2019

LOGICAL OPERATOR EXAMPLES


proposition operator examples.
logical operator examples
                                           

                                           Also search for this
                                     What is proposition?   click it
                        What is of logical operators define types?   click it






     1.        Let p, q, and r be the propositions
               p: You have the flu.
               q: You miss the final examination.
                r: You pass the course.

            a.       p ® q
If you have the flu, then you will miss the final examination.
            b.      Ø q « r 
You will not miss the final examination if and only if you pass the course.
            c.       q ® Ør
If you miss the final examination, then you will fail the course.
            d.      p v q V r
You have the flu either you miss the examination, or you pass the course.
             e.      (p ® Ør) v (q ® Ør)
If you have the flu, then you will not pass the course, or if you miss the final examination, then you will fail the course
            f.        (p /\ q) V (Øq /\ r)
You have the flu and you miss the final examination, or you will not miss the final examination and you pass the course.

     2.       Let p, q, and r be the propositions 
            p: You get an A on the final exam.
            q: You do every exercise in this book.
            r: You get an A in this class.
Write these propositions using p, q, and r and logical connectives.

a.       You get an A in this class, but you do not do every exercise in this book.
r Ù Øq
b.      You get an A on the final, you do every exercise in this book, and you get an A in this class.
p Ù q Ù r
c.       To get an A in this class, it is necessary for you to get an A on the final.
r ® p
d.      You get an A on the final, but you don't do every exercise in this book; nevertheless, you get an A in this class.
p Ù Øq Ù r
e.      Getting an A on the final and doing every exercise in this book is sufficient for getting an A in this class.
(p Ù q) ® r
f.        You will get an A in this class if and only if you either do every exercise in this book or you get an A on the final.
r « (q Ú p)

     3.       Construct a truth table for each of these compound propositions.
a.       p ®(Øq v r)

p
q
r
Ø q
Ø q Ú r
p ® (Ø q Ú r)
T
T
T
F
T
T
T
T
F
F
F
F
T
F
T
T
T
T
T
F
F
T
T
T
F
T
T
F
T
T
F
T
F
F
F
T
F
F
T
T
T
T
F
F
F
T
T
T

b.      Ø p ® (q ® r)

p
q
r
Ø p
q ® r
Øp ® ( q®r)
T
T
T
F
T
T
T
T
F
F
F
T
T
F
T
F
T
T
T
F
F
F
T
T
F
T
T
T
T
T
F
T
F
T
F
F
F
F
T
T
T
T
F
F
F
T
T
T

c.       (p ® q) v (ØP ® r)

p
q
r
Ø p
p ® q
Øp ® r
(p ® q) Ú (Øp ® r)
T
T
T
F
T
T
T
T
T
F
F
T
T
T
T
F
T
F
F
T
T
T
F
F
F
F
T
T
F
T
T
T
T
T
T
F
T
F
T
T
F
T
F
F
T
T
T
T
T
F
F
F
T
T
F
T

d.   (p ® q) Ù ( Ø p ® r)

p
q
r
Ø p
p ® q
Øp ® r
(p ® q) Ù (Øp ® r)
T
T
T
F
T
T
T
T
T
F
F
T
T
T
T
F
T
F
F
T
F
T
F
F
F
F
T
F
F
T
T
T
T
T
T
F
T
F
T
T
F
F
F
F
T
T
T
T
T
F
F
F
T
T
F
F

e.      (p « q)v( Øq « r)

p
q
r
Ø q
p « q
Øq « r
(p « q) Ú (Øq « r)
T
T
T
F
T
F
T
T
T
F
F
T
T
T
T
F
T
T
F
T
T
T
F
F
T
F
F
F
F
T
T
F
F
F
F
F
T
F
F
F
T
T
F
F
T
T
T
T
T
F
F
F
T
T
F
T

f.        ( Ø p « Øq) « (q«r)

p
q
r
Øp
Ø q
Øp « Øq
q « r
(Øp « Øq) « (q « r)
T
T
T
F
F
T
T
T
T
T
F
F
F
T
F
F
T
F
T
F
T
F
F
T
T
F
F
F
T
F
T
F
F
T
T
T
F
F
T
F
F
T
F
T
F
F
F
T
F
F
T
T
T
T
F
F
F
F
F
T
T
T
T
T

     4.        Evaluate each of these expressions.

a.       I 1000  Ù (0 1011 Ú I 1011)
0 1011  Ú  I 1011 = 11011
I 1000   Ù 11011 = 11000

b.      (0 IIII Ù I 0101)  Ú 0 1000  
0 IIII Ù I 0101 = 00101
00101  Ú 0 1000 = 01101

c.       (01010 Å 11011) Å 0 1000   
01010 Å 11011 = 10001
10001 Å 0 1000   = 11001

d.      (11011 Ú 0 1010) Ù (10001 Ú 11011)
11011 Ú 0 1010 = 11011
10001 Ú 11011 = 11011
11011 Ù 11011 = 11011

                                                                    MATH LABEL