Simplification of Boolean Expressions

In this video lecture we have simplified some Boolean expressions with the help of Boolean laws and Boolean properties. We have also used De Morgan theorem to simplify these questions.
Laws of Boolean algebra ----
• Laws of Boolean Algebr...
Logic circuit for boolean expressions ----
• Logic Circuit of Boole...
De Morgan's Theorem ----
• De Morgan's Theorems |...

[@halfheadcrusher4302]

Thank You Sir You saved my exam thanks once again you teach better than my school teacher

[@anjanakrishnan2073]

I had a doubt in the same sum where my teacher solved it in a much complicative way , this video was indeed helpful sir!!....Thank you sir!!👍👍

[@abhisheksinghbhau5797]

3rd question not done rightly.

[@uperdraporwalrinkuporwal6514]

Thanks sir 👍👍 Very helpful video

[@MohitVerma-mn9yd]

Fabb...... Bhai..

[@Larry21924]

This is beyond brilliant. A comparable book I read rewired my thought process. "A Life Unplugged: Reclaiming Reality in a Digital Age" by Theodore Blaze

[@steinsr3819]

Thank you sir it's very useful for me💖

[@manasgupta5004]

Thx sir , Much helpful! Greetings

[@journey844]

really very helpful...

[@coderguy339]

very very helpful ! thanx a ton!!!

[@harshsaini6741]

Easy explanation sir ji👍👍

[@gaudiumha9003]

Thanks so much sooooo useful ♡♡♡♡

[@RERE-gv2sk]

Thanks !!

[@saquibkhan9751]

Masha allah

[@deckerwellington6163]

Please why is A complement + C complement in (question 3) equal to 1? How?

[@Nitin_03_pawar]

O apne aap aya

[@notnemeziz]

kyuki 1+ anything=1......as 1 is the max value u can get

[@cr7__edits0]

I understand sir ❤

[@nageensabrimathstutorial763]

ABC[AB+C' (BC+AC)] solve this please

[@studyroomtargetssc]

Good

[@jeelezindagi7664]

(X+Y)(X bar +Y+Z) Can u please tell me the solution of this please sir please

[@shubhrasrivastava5586]

Take Y common Y+[x(x'+z)] Than open bracket by taking x inside Y+(xx'+xz) Y+(0+xz) Y+(xz) Yx+yz

[@prokwill3025]

Y+XZ

[@kilarisrinivas4083]

1:35 how ?? Someone explain

[@devilfrost9491]

He took A common and in bracket it become 1 + B' A(1 +B)' = A +AB'

[@lunchbirds9040]

AB+C(D+E) sir please solve this question

[@ramachandraiahramachandra2406]

I think it is not possible to reduce further

[@user-du1fy9tk8j]

If A+(A.B) =A[1+B'] But A+(AB)=A+B. ?

[@TheLearnersChannel]

Sameer, actually according to boolean laws any number added to 1 always gives 1. So A+AB = A.1 + A.B = A( 1+B ) = A. 1 = A

[@saqibqureshi4317]

#The Learners Channel AB'CAB then how did you do this whole term zero B'.B = 1 and A.A = A not zero their is not A.A'

[@TheLearnersChannel]

B'B = A'A = C'C = 0. Don't get confuse with the variables. Product or any variable with its component is always zero.

[@lunchbirds9040]

xy'+x'y please help me also solve please sir

[@TheLearnersChannel]

Entertainment & education.......x'y + xy' is the expansion of X + Y where + is exclusive addition which means exclusive addition of x & y will give x'y + xy'