This document contains information about Md. Ikbol Hossain including his educational and professional background. It lists his current position as Lecturer (ICT) at Adamjee Cantonment College and previous position as Ex-Lecturer (ICT) at BAF Shaheen College Dhaka. It also provides his contact information and summaries some theorems and proofs related to Boolean algebra.

1. MD. IKBAL HOSSAIN
LECTURER (ICT)
ADAMJEE CANTONMENT COLLEGE
1.EX. LECTURER (ICT)
BAF SHAHEEN COLLEGE DHAKA
2.BAF SHAHEEN COLLEGE
SHAMSHERNAGAR
B.Sc & M.Sc (1st class)
APPLIED PHYSICS AND
ELECTRONIC ENGINEERING.
RAJSHAHI UNIVERSITY.
Cell: 01734-580534
(imo,whatsapp,vaiber)
Email: ikbal.polua@gmail.com
঄ট ো- ঴োটজলনঃ
১। ভো঱ ছোটের ঴ক঱ ববশলশযিআ অশম ঄জজন
করব।অমোর পরীক্ষো ভো঱ ঵টব।঴ব শলক্ষক অমোটক
ভো঱বো঴টব।
২। অশম লশিলো঱ী মন ও দেট঵র ঄শিকোরী।জীবটন
঄বলিআ ঴ফ঱ ঵ব।
৩। অশম ঴ুস্থ থোকব,঴ুশি ঵ব।
৪। অশম ঴ো঵঴ী ও লশিমোন।শনটজর জনি ঴ুন্দর ভশব঳িৎ
গড়ব।
৫। অমোর কোজ করটে ভো঱ ঱োটগ;দ঱িোপড়ো একশ
কোজ,েোআ দ঱িোপড়ো করটেও ভো঱ ঱োটগ।অশম শনয়শমে
পড়োটলোনো কটর ভো঱ ছোে ঵ব,পরীক্ষোয় ও ভো঱ করব।
৬। দফ঴বুটক বো দমোবোআট঱ ঄যথো ঴ময় নয দকন
করব,অমোর দেো ঄টনক পড়ো বোশক।
৭। অমোর কথো বোেজো ও বিোব঵োটরর মোিিটম ঴বোর মন
জয় করব।অমোর কথোআ কয পোটব এমন কথো কোওটক
ব঱বনো।
৮। অশম দবশল জোশন এআ দভটব ঄঵ংকোর করব নো,যোরো
জোনটে চোআ েোটেরটক ঴ুন্দর ভোটব বুু্শঝটয় শেব।঴ব
বন্ধু রো অমোটক শনটয় গবজ করটব।
ছোেজীবটন ঴ফ঱েো ঄জজটনর উপোয়
১। ভো঳োর েক্ষেো ঄জজন।
২। দ঱িোপড়োয় দ঱টগ থোকো।
৩। বুটঝ পড়ো ও দ঱িো।
৪। ঴ক঱ শব঳টয় ঴মোন গুরুত্ত দেওয়ো।
৫। দ্রুে পড়োর ঄ভিো঴ করো।
৬। ভো঱ দনো ঴ংগ্র঵ করো ।
৭। আশেবোচক শচন্তো করো।
৮। রুশ নমোশফক জীবন পশরচো঱নো।
৯। শনটজ শনটজ পরীক্ষো দেওয়ো।
১০। স্বোটস্থির প্রশে যত্নবোন ঵ওয়ো।

2. HSC ICT Chaper 3:2nd
Part(Digital Device) BOOLEAN ALGEBRA
MD.IKBAL HOSSAIN,LECTURER(ICT),ADAMJEE CANTONMENT COLLEGE Page 1
Boolean Algebra
Proof of some Boolean Theorem
∴ 𝑳. 𝑯. 𝑺 = 𝑹. 𝑯. 𝑺(𝑷𝑹𝑶𝑽𝑬𝑫)
02. PROVE :A+A.B=A
L.H.S= A+A.B
=A.1+A.B [∵ 1. A = A]
=A(1+B)
=A.1 [∵ 1 + A = 1]
=A
∴ 𝑳. 𝑯. 𝑺 = 𝑹. 𝑯. 𝑺(𝑷𝑹𝑶𝑽𝑬𝑫)
03. PROVE: 𝐀(𝐀 + 𝐁) = 𝐀. 𝐁
L.H.S= A. (A + B)
=A. A+A.B
=0+A.B [∵ A. A = 0]
=A.B (R.H.S)
Basic Theorem
In Case Of Addition
01.A+0=A
02.A+1=1
03.A+A=A
04. A+𝐀=1
Basic Theorem
In Case Of Multiplication
01. A.0=0
02. A.1=A
03. A.A=A
04. A. 𝐀 =0
PROVE : A+𝐀=1
∴ 𝑳. 𝑯. 𝑺 = 𝑹. 𝑯. 𝑺(𝑷𝑹𝑶𝑽𝑬𝑫)
OR A+A
If A=1 then A = 0
=1+0
=1(R.H.S)
∴ 𝑳. 𝑯. 𝑺 = 𝑹. 𝑯. 𝑺(𝑷𝑹𝑶𝑽𝑬𝑫)
L.H.S= A+A
If A=0 then A = 1
=0+1
1 (R.H.S)
X-OR=A⨁B = A. B + A. B
X-NOR= 𝐀⨁𝐁 = 𝐀. 𝐁 + 𝐀. 𝐁
𝑫𝒆 𝑴𝒐𝒓𝒈𝒂𝒏′𝒔 𝒕𝒉𝒆𝒐𝒓𝒆𝒎:
𝟏. 𝐱 + 𝐲 = 𝐱. 𝐲
𝟐. 𝐱. 𝐲 = 𝐱 + 𝐲

3. HSC ICT Chaper 3:2nd
Part(Digital Device) BOOLEAN ALGEBRA
MD.IKBAL HOSSAIN,LECTURER(ICT),ADAMJEE CANTONMENT COLLEGE Page 2
∴ 𝑳. 𝑯. 𝑺 = 𝑹. 𝑯. 𝑺(𝑷𝑹𝑶𝑽𝑬𝑫)
04. PROVE :A+B.C=(A+B).(A+C)
L.H.S= A+B.C
=A.1+ B.C [∵ 1. A = A]
=A(1+B+C)+B.C [∵ 1 + 𝐴 = 1]
=A+A.B+A.C+B.C
=A.A+A.B+A.C+B.C [∵ A. A = A]
=A(A+B)+C(A+B)
=(A+B).(A+C) [R.H.S]
05.PROVE :(A+𝐀.B)=A+B
∴ 𝑳. 𝑯. 𝑺 = 𝑹. 𝑯. 𝑺(𝑷𝑹𝑶𝑽𝑬𝑫)
শবকল্প শনয়মঃ
L.H.S =(A+A.B)
(A+A). (A+B)
[∵A+B.C=(A+B).(A+C)]
=1.(A+B) [∵ 𝐀 + 𝐀 = 𝟏]
=A+B [R.H.S]
06.PROVE :( 𝐀+𝑨.𝐁)= 𝐀+𝐁
L.H.S =( A+A.B)
A.1+A.B
=A.(1+B)+ A. B [∵ 1 + A = 1]
=A+ A.B +A.B
=A+B.( A+A) [∵ A + A = 1]
=A+B.1 [1. A =A]
=A+B [R.H.S] ∴ 𝑳. 𝑯. 𝑺 = 𝑹. 𝑯. 𝑺(𝑷𝑹𝑶𝑽𝑬𝑫)
শবকল্প শনয়মঃ
L.H.S =( A+A.B)
=( A+A).( A+B) [∵ A + A = 1]
=1. ( A+B)
=A+B
=A+B [R.H.S]
∴ 𝑳. 𝑯. 𝑺 = 𝑹. 𝑯. 𝑺(𝑷𝑹𝑶𝑽𝑬𝑫)
L.H.S =(A+A.B)
A.1+A.B
=A.(1+B)+ A.B [∵ 1 + A = 1]
=A+A.B+A.B
=A+B.(A+A) [∵ A + A = 1]
=A+B.1 [1.A=A]
=A+B [R.H.S]
∴ 𝑳. 𝑯. 𝑺 = 𝑹. 𝑯. 𝑺(𝑷𝑹𝑶𝑽𝑬𝑫)
R.H.S=(A+B).(A+C)
= A.A+A.B+A.C+B.C
= A+A.B+A.C+B.C [∵ 𝐀. 𝐀 = 𝐀]
= A(1+B+C)+B.C
= A.1+ B.C [∵ 𝟏 + 𝐀 = 𝟏]
= A+B.C (L.H.S)

4. HSC ICT Chaper 3:2nd
Part(Digital Device) BOOLEAN ALGEBRA
MD.IKBAL HOSSAIN,LECTURER(ICT),ADAMJEE CANTONMENT COLLEGE Page 3
𝟎𝟕. 𝐏𝐑𝐎𝐕𝐄: 𝐀⨁𝐁 = 𝐀. 𝐁 + 𝐀. 𝐁
LHS=A⨁B
= A. B + A. B [∵ A⨁B = A. B + A. B]
=A. B. A. B [∵ X + Y = X. Y]
=(A+B).(A+B) [∵ X. Y = X + Y]
= (A+B).( A+B)
=A. A+A.B+A. B+B. B
=0+A.B+A. B+0 [∵ X. X = 0]
=A.B+A. B [R.H.S]∴ 𝐋. 𝐇. 𝐒 = 𝐑. 𝐇. 𝐒(𝐏𝐑𝐎𝐕𝐄𝐃)
𝟎𝟖. 𝑷𝑹𝑶𝑽𝑬: 𝑨⨁𝑩⨁𝑪 = 𝑨. 𝑩. 𝑪 + 𝑨. 𝑩. 𝑪 + 𝑨. 𝑩. 𝑪 + 𝑨. 𝑩. 𝑪
∴ 𝑳. 𝑯. 𝑺 = 𝑹. 𝑯. 𝑺(𝑷𝑹𝑶𝑽𝑬𝑫)
LHS= 𝐴⨁𝐵⨁𝐶
Let 𝐴⨁𝐵 = 𝑃
Then 𝐴⨁𝐵⨁𝐶 = 𝑃⨁𝐶
= 𝑃.C+ 𝑃. 𝐶 [∵ 𝐴⨁𝐵 = 𝐴. 𝐵 + 𝐴. 𝐵]
= 𝐴⨁𝐵. 𝐶 + 𝐴⨁𝐵. 𝐶
= ( 𝐴. 𝐵 + 𝐴. 𝐵 ). 𝐶 + ( 𝐴. 𝐵 + 𝐴. 𝐵). 𝐶 [∵ 𝐴⨁𝐵 = 𝐴. 𝐵 + 𝐴. 𝐵]
= 𝐴. 𝐵. 𝐶 + 𝐴. 𝐵. 𝐶 + 𝐴. 𝐵. 𝐶 + 𝐴. 𝐵. 𝐶
= 𝐴. 𝐵. 𝐶 + 𝐴. 𝐵. 𝐶 + 𝐴. 𝐵. 𝐶 + 𝐴. 𝐵. 𝐶 [R.H.S]
𝟎𝟗. 𝑷𝑹𝑶𝑽𝑬: 𝑨⨁𝑩⨁𝑪 = 𝑨. 𝑩. 𝑪 + 𝑨. 𝑩. 𝑪 + 𝑨. 𝑩. 𝑪 + 𝑨. 𝑩. 𝑪
∴ 𝑳. 𝑯. 𝑺 = 𝑹. 𝑯. 𝑺(𝑷𝑹𝑶𝑽𝑬𝑫)
LHS= 𝐴⨁𝐵⨁𝐶
Let 𝐴⨁𝐵 = 𝑃
Then 𝐴⨁𝐵⨁𝐶 = 𝑃⨁𝐶
=P.C+ 𝑃. 𝐶 [∵ 𝐴⨁𝐵 = 𝐴. 𝐵 + 𝐴. 𝐵]
= 𝐴⨁𝐵. 𝐶 + 𝐴⨁𝐵. 𝐶
= ( 𝐴. 𝐵 + 𝐴. 𝐵). 𝐶 +( A.B+ 𝐴. 𝐵 ). 𝐶 [∵ 𝐴⨁𝐵 = 𝐴. 𝐵 + 𝐴. 𝐵]
= 𝐴. 𝐵. 𝐶 + 𝐴. 𝐵. 𝐶 + 𝐴. 𝐵. 𝐶 + 𝐴. 𝐵. 𝐶 [R.H.S]

5. HSC ICT Chaper 3:2nd
Part(Digital Device) BOOLEAN ALGEBRA
MD.IKBAL HOSSAIN,LECTURER(ICT),ADAMJEE CANTONMENT COLLEGE Page 4
10. Simplify: 𝑨. 𝑩( 𝑨 + 𝑩)
=( 𝐴 + 𝐵)( 𝐴 + 𝐵)
= 𝐴. 𝐴 + 𝐴. 𝐵 + 𝐴. 𝐵 + 𝐵. 𝐵
= 𝐴 + 𝐴. 𝐵 + 𝐴. 𝐵
= 𝐴(1 + 𝐵 + 𝐵)
= 𝐴. 1
= 𝐴 (Ans)
11. Simplify: 𝑨 + 𝑩 + 𝑨
= 𝐴. 𝐵 + 𝐴
= 𝐴 + 𝐴. 𝐵
=(A+ 𝐴). (𝐴 + 𝐵)
=1. ( 𝐴 + 𝐵)
=(𝐴 + 𝐵)(Ans)
12.Simplify: A.B+ 𝑨. 𝑩 + 𝑨. 𝑩
=B.(A+ 𝐴) + A. 𝐵
=B.1+A. 𝐵
=B+A. 𝐵
=(B+A).(B+ 𝐵)
=(A+B).1
=(A+B) (Ans)
13.Simplify: 𝐀 + 𝐁 + 𝐂 + 𝐁. 𝐂
= 𝐴. 𝐵. 𝐶+B.C
= 𝐴. 𝐵. 𝐶+B.C
=C.( 𝐴. 𝐵+B)
=C.( 𝐴+B).( 𝐵 +B)
=C. ( 𝐴+B).1
=C. ( 𝐴+B) (Ans)
= 𝑋. 𝑌. (𝑍 + 𝑋)
14. Simplify: 𝑿 + 𝒀(𝒁 + 𝑿)
= 𝑋. (𝑌 + (𝑍 + 𝑋))
= 𝑋. (𝑌 + 𝑍. 𝑋)
= 𝑋. (𝑌 + 𝑍. 𝑋)
= 𝑋. 𝑌 + 𝑋. 𝑍. 𝑋
= 𝑋. 𝑌 + (𝑋. 𝑋). 𝑍
= 𝑋. 𝑌 + 0. 𝑍
= 𝑋. 𝑌+0
= 𝑋. 𝑌 (Ans)
15. Simplify: ( 𝐌 + 𝐍). ( 𝐌 + 𝐍)
= ( 𝑀 + 𝑁) + (𝑀 + 𝑁)
= 𝑀. 𝑁+ 𝑀. 𝑁
=M. 𝑁+ 𝑀.N
= 𝑀⨁𝑁 (Ans)

6. HSC ICT Chaper 3:2nd
Part(Digital Device) BOOLEAN ALGEBRA
MD.IKBAL HOSSAIN,LECTURER(ICT),ADAMJEE CANTONMENT COLLEGE Page 5
16. Simplify: (X+Y).( 𝐗+Z).(Y+Z)
=(X.X+X.Z+X. Y + Y. Z). (Y+Z)
=(0+ X.Z+X. Y + Y. Z). (Y+Z)
=(X.Z+X. Y + Y. Z). (Y+Z)
=X.Y.Z+X. Y. Y+Y. Y. Z+ X.Z.Z+X. Y. Z+ Y. Z.Z
= X.Y.Z+X. Y+ Y. Z+ X.Z+X. Y. Z+ Y. Z
= X.Y.Z+ Y. Z+X. Y. Z+ Y. Z+X. Y+ X.Z
= Y.Z(X+1+X+1) +X. Y+ X.Z
=Y.Z.1+X. Y+ X.Z
=Y.Z(X+X)+ X. Y+ X.Z
=X.Y.Z+X.Y.Z+ X. Y+ X.Z
=X.Y.Z+ X. Y+ X.Y.Z+ X.Z
=X.Y(Z+1)+X.Y(Z+1)
=X.Y+ X.Y (Ans)
∴ 𝐋. 𝐇. 𝐒 = 𝐑. 𝐇. 𝐒(𝐏𝐑𝐎𝐕𝐄𝐃)
17. PROVE: (X+Y).( 𝐗+Z).(Y+Z)= (X+Y).( 𝐗+Z)
L.H.S=(X+Y).( X+Z).(Y+Z)
=(X.X+X.Z+X. Y + Y. Z). (Y+Z)
=(0+ X.Z+X. Y + Y. Z). (Y+Z)
=(X.Z+X. Y + Y. Z). (Y+Z)
=X.Y.Z+X. Y. Y+Y. Y. Z+ X.Z.Z+X. Y. Z+ Y. Z.Z
= X.Y.Z+X. Y+ Y. Z+ X.Z+X. Y. Z+ Y. Z
= X.Y.Z+ Y. Z+X. Y. Z+ Y. Z+X. Y+ X.Z
= Y.Z(X+1+X+1) +X. Y+ X.Z
=Y.Z.1+X. Y+ X.Z
= Y.Z+X. Y+ X.Z
=0+ Y.Z+X. Y+ X.Z
= X.X+ Y.Z+X. Y+ X.Z
= X.X+X. Y+ X.Z+ Y.Z
=X. (X + Y) + Z(X + Y)
=(X+Y).( X +Z) [R.H.S]

7. HSC ICT Chaper 3:2nd
Part(Digital Device) BOOLEAN ALGEBRA
MD.IKBAL HOSSAIN,LECTURER(ICT),ADAMJEE CANTONMENT COLLEGE Page 6
18.Prove : X.Y+𝐗. 𝐙+ Y.Z= X.Y+𝐗. 𝐙
L.H.S= X.Y+X. Z+ Y.Z
= X.Y.1+X. Z. 1+ Y.Z.1
=X.Y(Z+Z)+X. Z(Y + Y)+Y.Z(X+X) [∵ A + A = 1]
=X.Y.Z+X.Y. Z+X. Z. Y + X. Z. Y + Y. Z. X + Y. Z. X.
=(X.Y.Z+ Y. Z. X )+ (X. Z. Y + Y. Z. X. ) +X. Z. Y+X.Y. Z
=(X.Y.Z+X.Y.Z)+( X.Y.Z+X.Y.Z)+ X. Z. Y+X.Y. Z
=X.Y.Z+X.Y.Z+X. Z. Y+X.Y. Z [A+A=A]
= X.Y.Z+X.Y. Z+X.Y.Z+X. Z. Y
=X.Y(Z+Z)+ X. Z(Y + Y)
=X.Y.1+X. Z. 1 [∵ A + A = 1]
= X.Y+X. Z [R.H.S] [∴ 𝑳. 𝑯. 𝑺 = 𝑹. 𝑯. 𝑺(𝑷𝑹𝑶𝑽𝑬𝑫)]
𝟏𝟗. 𝐈𝐟 𝐅 = 𝐗. 𝐘 + 𝐗. 𝐘. 𝐙 𝐭𝐡𝐞𝐧 𝐩𝐫𝐨𝐯𝐞 𝐭𝐡𝐚𝐭
( 𝐢). 𝐅. 𝐅 = 𝟎
(𝐢𝐢). 𝐅 + 𝐅 = 𝟏
F = X. Y + X. Y. Z
∴ F = Y. (X. Z)
Solve: given that,
=Y.( X + X. Z)
=Y.( X + X). (X + Z)
= Y.1. (X + Z)
= Y. (X. Z)
∴ F = Y. (X. Z) … … (1)
=Y + (X. Z)
=Y +X.Z
∴ F = Y +X.Z … … (2)
∴ L. H. S (i) F. F =
=Y. (X. Z). (Y +X.Z)
=Y. Y. (X. Z) + Y. (X. Z). (X. Z)
=0. (X. Z)+Y.0
=0+0
=0(R.H.S)
∴𝐋.𝐇.𝐒=𝐑.𝐇.𝐒(𝐏𝐑𝐎𝐕𝐄𝐃)
∴ 𝑳. 𝑯. 𝑺 ( 𝒊𝒊) 𝑭 + 𝑭 = =Y. ( 𝑿. 𝒁) + 𝒀 +X.Z
= 𝑌 +Y.( 𝑋. 𝑍) + (𝑋. 𝑍)
=( 𝑌+ Y).( 𝑌 + 𝑋. 𝑍)+ 𝑋. 𝑍
=1. ( 𝑌 + 𝑋. 𝑍)+ 𝑋. 𝑍
= 𝑌 + 𝑋. 𝑍+ 𝑋. 𝑍
= 𝑌 + 1
=1(R.H.S) ∴ 𝑳. 𝑯. 𝑺 = 𝑹. 𝑯. 𝑺(𝑷𝑹𝑶𝑽𝑬𝑫)

8. HSC ICT Chaper 3:2nd
Part(Digital Device) BOOLEAN ALGEBRA
MD.IKBAL HOSSAIN,LECTURER(ICT),ADAMJEE CANTONMENT COLLEGE Page 7
20.Simplify: A.B.C+A.𝐁. 𝐂 + 𝐀. 𝐁. 𝐂 + 𝐀. 𝐂
=A.B.C+A.B.C+ A.B. C+A. C
=A.B.C+ A.B. C+A. C
=A.C(B+B)+ A. C
=A.C.1+A. C
=A.C+A. C
=C(A+A)
=C.1
=C (Ans)
21.Simplify: (B+𝐂). ( 𝐁 + 𝐂) + 𝐀 + 𝐁 + 𝐂
=(B+ 𝐶). ( 𝐵 + 𝐶) + 𝐴. 𝐵. 𝐶
=B. 𝐵+B.C+ 𝐵. 𝐶 + 𝐶. 𝐶 + 𝐴. 𝐵. 𝐶
=0+ B.C+ 𝐵. 𝐶 + 0 + 𝐴. 𝐵. 𝐶
= B.C+ 𝐵. 𝐶 + 𝐴. 𝐵. 𝐶
=B.C+ 𝐵.( 𝐶 + 𝐴. 𝐶)
=B.C+ 𝐵. ( 𝐶 + 𝐴). ( 𝐶 + 𝐶)
=B.C+ 𝐵. ( 𝐶 + 𝐴). 1
=B.C+ 𝐵. 𝐶+A. 𝐵
= 𝐵⨁𝐶 + 𝐴. 𝐵 (Ans) [∵ 𝐴. 𝐵 + 𝐴. 𝐵 = 𝐴⨁𝐵]

9. HSC ICT Chaper 3:2nd
Part(Digital Device) BOOLEAN ALGEBRA
MD.IKBAL HOSSAIN,LECTURER(ICT),ADAMJEE CANTONMENT COLLEGE Page 8
𝟐𝟐. 𝐏𝐑𝐎𝐕𝐄: 𝐀. 𝐁. 𝐂 + 𝐀. 𝐁. 𝐂 + 𝐀. 𝐁. 𝐂 + 𝐀. 𝐁. 𝐂 = 𝐀⨁𝐁⨁𝐂
= ( 𝐴. 𝐵 + 𝐴. 𝐵 ). 𝐶 + ( 𝐴. 𝐵 + 𝐴. 𝐵). 𝐶
∴ 𝑳. 𝑯. 𝑺 = 𝑹. 𝑯. 𝑺(𝑷𝑹𝑶𝑽𝑬𝑫)
LHS= 𝐴. 𝐵. 𝐶 + 𝐴. 𝐵. 𝐶 + 𝐴. 𝐵. 𝐶 + 𝐴. 𝐵. 𝐶
= 𝐴. 𝐵. 𝐶 + 𝐴. 𝐵. 𝐶 + 𝐴. 𝐵. 𝐶 + 𝐴. 𝐵. 𝐶
= 𝐴⨁𝐵. 𝐶 + 𝐴⨁𝐵. 𝐶 [∵ 𝐴⨁𝐵 = 𝐴. 𝐵 + 𝐴. 𝐵][∵ 𝐴⨁𝐵 = 𝐴. 𝐵 + 𝐴. 𝐵]
Let 𝐴⨁𝐵 = 𝑃
Then 𝐴⨁𝐵. 𝐶 + 𝐴⨁𝐵. 𝐶 = 𝑃.C+ 𝑃. 𝐶
= 𝑃.C+ 𝑃. 𝐶
= 𝑃⨁𝐶[Putting the value of P]
= 𝐴⨁𝐵⨁𝐶 [R.H.S]
𝟐𝟑. 𝐏𝐑𝐎𝐕𝐄: 𝐀. 𝐁. 𝐂 + 𝐀. 𝐁. 𝐂 + 𝐀. 𝐁. 𝐂 + 𝐀. 𝐁. 𝐂 = 𝐀⨁𝐁⨁𝐂
∴ 𝑳. 𝑯. 𝑺 = 𝑹. 𝑯. 𝑺(𝑷𝑹𝑶𝑽𝑬𝑫)
L.H.S= 𝐴. 𝐵. 𝐶 + 𝐴. 𝐵. 𝐶 + 𝐴. 𝐵. 𝐶 + 𝐴. 𝐵. 𝐶
= ( 𝐴. 𝐵 + 𝐴. 𝐵). 𝐶 +( A.B+ 𝐴. 𝐵 ). 𝐶
= 𝐴⨁𝐵. 𝐶 + 𝐴⨁𝐵. 𝐶 [∵ 𝐴⨁𝐵 = 𝐴. 𝐵 + 𝐴. 𝐵] [∵ 𝐴⨁𝐵 = 𝐴. 𝐵 + 𝐴. 𝐵]
Let 𝐴⨁𝐵 = 𝑃
Then 𝐴⨁𝐵. 𝐶 + 𝐴⨁𝐵. 𝐶 =P.C+ 𝑃. 𝐶
= 𝑃⨁𝐶 [∵ 𝐴⨁𝐵 = 𝐴. 𝐵 + 𝐴. 𝐵]
= 𝐴⨁𝐵⨁𝐶 [R.H.S] [Putting the value of P]
𝟐𝟒. 𝐏𝐫𝐨𝐯𝐞 𝐭𝐡𝐚𝐭: 𝐀. 𝐁. 𝐂 + 𝐀. 𝐁. 𝐂 + 𝐀. 𝐁. 𝐂 + 𝐀. 𝐁. 𝐂 = 𝐀. 𝐁 + 𝐁. 𝐂 + 𝐀. 𝐂
L.H.S= 𝐴. 𝐵. 𝐶 + 𝐴. 𝐵. 𝐶 + 𝐴. 𝐵. 𝐶 + 𝐴. 𝐵. 𝐶
= 𝐴. 𝐵. 𝐶 + 𝐴. 𝐵. 𝐶 + 𝐴. 𝐵. 𝐶 + 𝐴. 𝐵. 𝐶 + 𝐴. 𝐵. 𝐶 + 𝐴. 𝐵. 𝐶 [∵ 𝒂 + 𝒂 + 𝒂 = 𝒂]
= 𝐴. 𝐵. 𝐶 + 𝐴. 𝐵. 𝐶 + 𝐴. 𝐵. 𝐶 + 𝐴. 𝐵. 𝐶 + 𝐴. 𝐵. 𝐶 + 𝐴. 𝐵. 𝐶
= 𝐵. 𝐶( 𝐴 + 𝐴) + 𝐴. 𝐶( 𝐵 + 𝐵) + 𝐴. 𝐵( 𝐶 + 𝐶)
=B.C.1+A.C.1+A.B.1 ∵ 𝑿 + 𝑿 = 𝟏
=B.C+A.C+A.B
=A.B+B.C+A.C(R.H.S) ∴ 𝑳. 𝑯. 𝑺 = 𝑹. 𝑯. 𝑺(𝑷𝑹𝑶𝑽𝑬𝑫)

10. HSC ICT Chaper 3:2nd
Part(Digital Device) BOOLEAN ALGEBRA
MD.IKBAL HOSSAIN,LECTURER(ICT),ADAMJEE CANTONMENT COLLEGE Page 9
25.Prove that:
( 𝐀 + 𝐁). (𝐀 + 𝐁 )=0
L.H.S= ( 𝐴 + 𝐵). (𝐴 + 𝐵 )
=(𝐴. 𝐵). (𝐴.𝐵) [∵ ( 𝐱 + 𝐲) = 𝐱. 𝐲]
=( 𝐴. 𝐵). ( 𝐴. 𝐵)
=( 𝐴. 𝐴). ( 𝐵. 𝐵) [∵ 𝑿. 𝑿 = 𝟎]
=0.0
=0(R.H.S) ∴ 𝑳. 𝑯. 𝑺 = 𝑹. 𝑯. 𝑺(𝑷𝑹𝑶𝑽𝑬𝑫)
26.Prove that:
𝐀 + 𝐀. 𝐁 + 𝐀. 𝐁 =1
L.H.S= 𝐴 + 𝐴. 𝐵 + 𝐴. 𝐵
=A+ 𝐴( 𝐵 + 𝐵) [∵ 𝑿 + 𝑿 = 𝟏]
=A+ 𝐴. 1
=A+ 𝐴 [∵ 𝑿 + 𝑿 = 𝟏]
=1 (R.H.S)
27.Prove that :
𝐑. 𝐒. 𝐓.(𝐑 + 𝐒 + 𝐓)=𝐑. 𝐒. 𝐓
L.H.S=(𝑅+ 𝑆+ 𝑇).( 𝑅. 𝑆. 𝑇)
= 𝑅. ( 𝑅. 𝑆. 𝑇)+ 𝑆 .( 𝑅. 𝑆. 𝑇) + 𝑇. (𝑅. 𝑆. 𝑇)
=(𝑅. 𝑅. 𝑆. 𝑇)+ (𝑅. 𝑆. 𝑆. 𝑇)+ (𝑅. 𝑆. 𝑇. 𝑇)
= 𝑅. 𝑆. 𝑇 + 𝑅. 𝑆. 𝑇 + 𝑅. 𝑆. 𝑇
= 𝑅. 𝑆. 𝑇(1 + 1 + 1)
= 𝑅. 𝑆. 𝑇. 1
= 𝑅. 𝑆. 𝑇(R.H.S) ∴ 𝑳. 𝑯. 𝑺 = 𝑹. 𝑯. 𝑺(𝑷𝑹𝑶𝑽𝑬𝑫)
28. simplify: 𝑨. 𝑩. 𝑪. 𝑫
=𝐴. 𝐵. 𝐶 + 𝐷
=𝐴. 𝐵. 𝐶 + 𝐷
=( 𝐴 + 𝐵). 𝐶 + 𝐷 (𝐴𝑛𝑠)
𝟐𝟗. 𝐒𝐢𝐦𝐩𝐥𝐢𝐟𝐲 ∶ ( 𝐀 + 𝐁). 𝐁. 𝐂
=(𝐴 + 𝐵). (𝐵. 𝐶) [∵ 𝐴. 𝐵 = 𝐴 + 𝐵]
=(𝐴 + 𝐵)(𝐵 + 𝐶) [∵ 𝐴 + 𝐵 = 𝐴. 𝐵]
=(𝐴 + 𝐵)(𝐵 + 𝐶) [∵ 𝐴 = 𝐴]
=( 𝐵 + 𝐴). ( 𝐵 + 𝐶)
=𝐵 + 𝐴. 𝐶 (Ans) [∵ 𝐴 + 𝐵. 𝐶 = (𝐴 + 𝐵). ( 𝐴 + 𝐶)]

11. HSC ICT Chaper 3:2nd
Part(Digital Device) BOOLEAN ALGEBRA
MD.IKBAL HOSSAIN,LECTURER(ICT),ADAMJEE CANTONMENT COLLEGE Page 10
30. Simplify: ( 𝑨. 𝑩 + 𝑨. 𝑩). ( 𝑨 + 𝑩)
= 𝐴. 𝐵. 𝐴. 𝐵. ( 𝐴 + 𝐵)
= 𝐴 + 𝐵 . 𝐴 + 𝐵 . ( 𝐴 + 𝐵)
=(𝐴 + 𝐵). ( 𝐴 + 𝐵). (𝐴 + 𝐵)
=( 𝐴 + 𝐵). ( 𝐴. 𝐴 + 𝐴. 𝐵 + 𝐴. 𝐵 + 𝐵. 𝐵)
=( 𝐴 + 𝐵). ( 𝐴. +𝐴. 𝐵 + 𝐴. 𝐵 + 0)
=( 𝐴 + 𝐵). ( 𝐴(1 + 𝐵 + 𝐵) + 0)
=( 𝐴 + 𝐵). 𝐴. (1 + 1)
=( 𝐴 + 𝐵). 𝐴. 1
=( 𝐴 + 𝐵). 𝐴
= 𝐴. 𝐴 + 𝐴. 𝐵
=0+A.B
=A.B (Ans)