Simplification of Boolean Expression using Boolean Algebra Rules | Important Question 3

[@dev77cmd]

One night before the exam, you cleared my mind about simplification of boolean expressions, thank you!!!

[@hamismohammed4851]

Am new her but I've learned something that I've been trying so hard to understand.. Now I got it..
I appreciate alot

[@M_Nagesh_kaushik]

Dil se respect h apke liye ❤

[@M_Nagesh_kaushik]

Sir bahut badiya samjhate ho❤❤❤

[@sankeertgandoli4368]

Helped me learn this concept in an easier way, thank you

[@talhabosstj1239]

Logic and rules are defined very finely

[@aminmousavi7813]

thankyou sir we need people like you

[@bdnr8346]

Namaskar Ji, Very good initiative

[@mkandawirehaggaipurpose3772]

This was just wonderful..... I appreciate the knowledge

[@ENGINEERINGTUTORIAL2468]

Thank you for your feedback.

[@JadHaddad-i7w]

Thank you so much, this shit made no sense to me until god sent your beautiful self my way

[@kamaraernestben4802]

Thank you very much I need more videos on boolean expression reduce method

[@anoopkrishnans6016]

Good sir🤗🤗

[@wrestlingfan8112]

Thank you! Great explanation! :)

[@jessamaeopondarulona4863]

Can you provide the link of your tutorial/explanation about the De Morgan's Theorem? Thank you.

[@teenasaini3553]

this is very good i understand it easily

[@vinodhar6620]

Great explanation.It helped me a lot.Thank you sir

[@aliciagenson2225]

thank you bbg that really helped

[@shaflaakter366]

Very nice class

[@toluwanidaniel7841]

Great explanation... Thanks man

[@LeBateauRouge]

why not doing D('not'ABC+1) instead of 'not'abc(1+D)?
thanks for answering in advance!

[@AnkiitOp]

Because d is not common in both terms bro

[@LightYagami-kira.]

Thank u very much sir🙏

[@hidajetkucevic9004]

very GOOD!

[@lishyyy2624]

Good video ❤

[@quizychamu717]

Tq u soooo much sir great work

[@talhabosstj1239]

Good points

[@ammusharmila9841]

Hi sir can you please simplify the following term. ab'd'+acd'+a'bc'd'+abc'd

[@zhar_eng8765]

A(D'+BC'D)

[@fifid7660]

Why did you add (1+D)?

[@unknowngeezer5029]

cant you just do absorption law straight away so a[b+c(ab+ac)] so b+ab = b so it final becomes a[b+c]

[@makeingoogle4829]

Thanks yoooouu also🙂🙂🙂

[@arpitranafromchandigarh]

Then what we can write if there is (ABC)' ?

[@timskirwa6034]

Very good

[@preetamsthings4189]

Thanks ☺️

[@ibrahimferoz1449]

Thanks really

[@bhagyashreemasalakitchen3673]

Thanks bro for your nice explanation 😇😇

[@parangnenengbgaming5277]

sorry how did u get the 1 in 2:00

[@gabrielbarrera3447]

Yeah idk either

[@gabrielbarrera3447]

But in that step yoy can apply the absorption rule. X+Xy=X. So B+ABC is just B. B representing X, y representing AC.

[@pratikchakraborty6354]

@@gabrielbarrera3447 he took B common from the first two terms (B +ABC) so obviously the expression is B(1+ABC) since there can't be 0 there.

[@jayaverma3940]

Thanks sir

[@rasalmba7328]

Yhe one kaise aa raha h

[@rudreshsingh4681]

Great Content Sir :)
Really helped me .

[@ENGINEERINGTUTORIAL2468]

Thanks for the feedback.

[@un9339]

Tq sir. it's really helpful to me

[@rehanahamad8977]

Any easy method sir

[@068fahimaakter3]

Tnx

[@amarzais804]

B+AB'+AB'C+A'B+AB'CD+A'BC'DE'
HOW TO SIMPLIFY

[@krishnanshgupta9827]

Rearrange
B+A'B+AB'+AB'CD+A'BC'DE'
B(1+A')+AB'(1+CD)+A'BC'DE'
B + AB' + A'BC'DE'
Taking only B + AB' --> B + A
B + A + A'BC'DE'
A + B(1+A'C'DE')
(A + B)
:)

[@talhabosstj1239]

Points

[@shivaniyadav7297]

U did it wrong in 1st question 1+x = x

[@charlesakintunde4767]

No it's 1

[@hadiamisbah3739]

@@charlesakintunde4767 how??
its x

[@abdullahbaig8700]

Its called the Dominance Law
A+1=1
Any Boolean Expression directly added into 1 will result in 1.
A.0=0 is also another expression.

[@yashthorve8130]

Women ☕

[@KABKAKIy_KAPATEJI_DOP]

Bruh, its called Identity Rule in OR-Form , and it has result = 1; A + 1= 1
In AND-Form it is; A × 1 = A