Redundancy Theorem (Boolean Algebra Trick)

Digital Electronics: Redundancy Theorem - A Boolean Algebra Trick
1) Requirements to apply the redundancy theorem.
2) Applications of redundancy theorem.
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[@vidhanmaheshwari2082]

This video is 5 years old but still is the best..Thanks a lot Sir.

[@vishaalkushwaha]

OLD IS GOLD💓🔥

[@gamar1226]

thank you !

[@vtuboi]

4th example Answer. G=(A+B)•(B'+C) Please correct me if I'm wrong

[@satyam1543]

@@vtuboi correct

[@vtuboi]

@@satyam1543 thank you 😊

[@gixr9421]

8 years now... Unbelievable.. True talent never gets old

[@tharushasanjaya7124]

😮 now this video is 9 years old. whatever it's the best yet

[@rohitnamballa236]

u r the best...5 years old video but still no other lecture is able to beat urs till now

[@Saurabh_91]

This video is 7 years old but there is no any videos on KZbin like that.... Thanks a lot ❤️

[@snehashishbanerjee2575]

🔴HW Example 2: ans:: F = AB' + BC 🔴HW Example 4: ans:: G = (A+B).(B'+C) [using Redundancy theorem]

[@gojosatoru2005]

thank god im right😄

[@amaansheikh6046]

@@gojosatoru2005 bca kr rhe ho kya

[@gojosatoru2005]

@@amaansheikh6046 what 😅 can you type in English

[@cryptoparadoxx]

it's wrong ex4 should be AB' +BC

[@varshaa14]

@@cryptoparadoxx yeah when you simplify it you get that as the answer

[@chido-in-the-lab1811]

proof of HW F= AB' + BC +AC F= AB' +BC + AC. 1 = AB'+ BC + AC(B+ B') =AB' + BC+ ACB + ACB' = AB'(1+C) + BC(1+A) = AB' + BC

[@sunilkumar-ls2yb]

hey i m unable to understand plz help

[@chido-in-the-lab1811]

@@sunilkumar-ls2yb . multiply AC by 1 because it's not associated with B or its compliment. . 1 is the same as B+B' according to boolean algebra. .use the distributive law . factor out the common factors . then use boolean algebra to simplify... Anything or'd with 1 is simply 1.

[@harshmk1052]

@@sunilkumar-ls2yb Seedhi baat hai - jo term ko hatana hai , jaise in this example we have to remove AC toh AC mein woh literal ka multiply karo jo complement hua hai . Matlab AC(B+B') Aise hi redundancy theorem ke dusre bhi question ke proof aa jayenge .

[@sunilkumar-ls2yb]

@@harshmk1052 thnx bro i understand that

[@pankajverma5749]

@@sunilkumar-ls2yb F=AB'+BC+AC F=AB'+BC+AC.1 F=AB'+BC+AC.(B+B') using distributive law F=AB'+BC+(ABC+AB'C) F=[AB'+AB'C]+ABC+BC F=AB'(1+C)+BC(1+A) F=AB'C+ABC F=AB'+AC

[@RahulMadhavan]

Solution for Ex 3: F = (A + B).(A' + C).(B + C) = (A + B).(A' + C).(B + C + AA') = (A + B).(A' + C).(B + C + A').(B + C + A) Rearranging, we get F = (A + B).(A + B + C).(A' + C).(A' + C + B) = ((A + B)(1 + C)).((A' + C)(1 + B)) = (A + B).(A' + C)

[@ekramulreza]

I got, F=(A+B)(B'+C) 🤔🤔🤔

[@RahulMadhavan]

@@ekramulreza Not possible. Given F = (A + B).(A' + C).(B + C). For A=0,B=1,C=0, F is true Now consider F'=(A+B)(B'+C). Then for A=0,B=1,C=0, F' is false.

[@bdrfuad]

there is an error in your solution of the distributive law.

[@RahulMadhavan]

@@bdrfuad typically one may ask "how did you get step ... in your solution". Which step do you have an issue with?

[@Dozby]

@@RahulMadhavan Chill, your answer is right. As well as his. The problem is (I guess) that he was solving problem number 4 [which is F=(A+B)(B'+C)] , not 3. [which is (A + B)(A' + C)] If I didn't do some calculation mistake, but I hope not... 😅

[@mrinmoypal9630]

You really help a lot to increase our interest in the subject.Every second of the video is of worth watching.Do continue your good work.It would be of great help if you come up with a series of electrical Machines.

[@renuchavre1818]

H.W. AB'+BC+AC Ans.By using redundancy theorem we can solve this problem: Rule:One variable is complemented. And here:B is complemented. So we have to omit the variables which doesn't contain even single complement.But we can't omit "BC" because as per the rule B is complemented and BC variable contain B. So we have to omit AC variable. And hence the answer is AB'+BC.

[@manaklalkanel1308]

Ex. 2 - AB`+BC And Ex.-4- (B`+c)(A+B)

[@rajeshsonga4404]

I can't do home work actually I am not getting them

[@PraveenPitchuka]

I don't do home work bc I'm homeless

[@dharmeshparmani8473]

@@rajeshsonga4404 In H.W. Problem , that is , question no. 2 , AC term is redundant there because it does not contain the complemented variable . In this question Variable "B" is complemented variable . So the answer is AB'+BC . Why becoz if you multiply "BC" by 1 then by Boolean Algebra there will be no change in it . As variable "A" multiplied by "1" gives Variable "A" itself . For example , take "A" = 1 and "1 AND 1" will give "1" and now take "A" = 0 and "0 AND 1" gives "0" only . So anyway we get "A" only as output . So Now "BC" multiplied by "1" is equivalent to BC(A+A') and then after opening the bracket and writing all the terms and then taking AB' and BC common , we will get AB'+BC as answer . H.W. Q 4 - (B'+C)( A+B) .

[@dharmeshparmani8473]

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[@koi1440]

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Had to watch a few parts again, doing project and biography on George Boole and describing/explaining how his mathematical formulas worked. Thank you for the help though, I needed it to better understand how this worked.

[@ibrahnoor6741]

Sir , your video is really helpful for my 1st semester exam.... please make some videos on digital electronics practical of Boolean algebra using breadboards , multimeter , ICs ,wires , etc .

[@gabrielebiscetti4286]

2) F = AB' + BC + AC Consensus between AB' and BC return AC. So AC is an prime implicant. In general, if "P" is an boolean polynomial and "I" is prime implicant, then P = P + I So AB' + BC = AB' + BC + AC 4) Boolean algebra is dual, so is the same procure of (2) but with OR, AND switched G = (A + B) (B' + C) Is it correct ?

[@harshachallagulla4321]

Yes

[@mohitkrishnayadav8302]

correct

[@bagadishashikiran2129]

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[@llllllllll5119]

7 years old but it still does the trick. You're the best.

[@owonubijobsunday4764]

I've been looking for this theorem for a while. Didn't know it exists. Thanks very much. Truth table proves it ☑

[@buseduban2272]

can you share how you prove it with the truth table please?

[@sidra3210]

Q.2 : F = AB' + BC + AC Answer : F = AB' + BC Q.4 : G = (A + B ) . ( B' + C ) . ( A + C ) Answer : G = (A + B ) . ( B' + C) Please this is connect?

[@019_deepsikhadas3]

yes

[@sidra3210]

@@019_deepsikhadas3 shukria✌

[@pratikkarale8461]

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[@michaelkaufman5119]

This explanation is much better than my university professor. Thank you for this.

[@canoselshekinah96]

Thank you, sir! Your lessons are easy to understand and are well explained.

[@rakshithkumar9381]

Thanks for the Video Sir,This helped a lot in Lockdown

[@aditivarshney3466]

FOR ii) AB'+BC=F for iv) G=(A+B).(B'+C)

[@top10decode67]

watching it in 2023 really makes lots of sense

[@nuvvin]

almost a decade now but still very useful

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[@ASocial093]

3rd rule says one variable is complemented but in (H.W) 2 i.e. [(A+B).(B'+C).(A'+C)] have two variable as complemented. Hence we cannot use Redundancy theorem. Also (H.W) 1 i.e. [A.B'+B.C+A.C] answer will be [A.B'+B.C]

[@hareramkumar8638]

this video is 7 year old but still is the best on internet..... thanks

[@erswatimaurya4228]

1. simplify the Boolean expression using Boolean algebra:- a) AB+AC+BC=AB+AC B) (A'+BC)' (A'B'+ABC) 2. State De-morgans law prove the following using Boolean algebra theorem:- AB+AC+BC=AB+AC

[@ajayakumarpradhan7408]

Swati Mourya how to solve it

[@sidhhagang226]

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[@barrykinnan5062]

right? ...like a min or two on how he uses the dist. propety

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[@siddharthtrivedi8586]

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[@chandanrayece]

2. AB'+BC 4.(A+B)(B'+C)

[@AbhaySingh-dw1qw]

Thank you sir 😍😍😍

[@daliamohamed5525]

ii) AB'+BC iv) (A+B).(B'+C)

[@STRANGER8049]

seeing this video after 6 years but now this video is best thanks so much sir now I am confident that I'll pass my exam

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I can't believe it's 8 years old and still all the concepts are same and so helpfull and education system is as it is

[@BlackOpal-dk6iv]

SIR I HAVE COMPLETED LECTURE 11 OUT OF 202. TREMENDOUS LECTURE AND THANKYOU FOR YOUR GUIDANCE !!!

[@KCEEE_VIPULKUMAR]

These videos are very perfect to understand Boolean Algebra Trick

[@harshiniadusumilli2509]

first homework question solution F=AB'+BC+AC F=AB'+BC+AC(B+B') F=AB'+BC+ABC+AB'C F=AB'(1+C)+BC(1+A) F=AB'+BC second homework question solution G=(A+B)(B'+C)(A+C)=(A+B)(B'+C) sir, will u please explain the second homework problem

[@anujsinghsomvanshi3385]

very nice trick sir. thank you so much.

[@thedelhiguy1112]

Using Redundancy Theorem We get :- F= A'B + BC. G = (A+B) . (B'+C).

[@im_kamalpal]

Thank u a lot Sir ❤️ Videos are 7 yrs old, but till now no good teacher as well as him.

[@brainhax6758]

Redunducy thrm can applied one more time in 3rd example (A+B)(A'+C) = A.A' +AC+A'B +BC=AC+A'B Correct me if am wrong?

[@ashish1284]

Correct

[@adidebdas2703]

understood concepts within minutes , thank you sir !

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your videos are really really good,i really like them,thank you for your work

[@DeyvesshKumar]

+Neso Academy Sir, can you solve these example I am stuck..Example 2, the AB i have done but BC' I am not able to solve. I have also taken (B'+B) But Nothing happened...

[@learningisanartandfree_mmr4888]

do any one solve this problem??????????

[@akshatjain4867]

how to solve this

[@hishmaalijinnah1987]

thank you very much for helping me with my doubt

[@meenakshi1327]

Sir , you are doing GOD'S WORK 😭😭😭THANKYOU SO FVKING MUCH

[@naughtyjiyansh533]

Very very nice explanation 👌 thanks for sharing

[@sanjaygovinds5014]

Thanks for the video it helped me a lot.

[@arnabdas7485]

For 2nd question answer is AB'+BC For 4th question answer is (A+B). (B'+C)

[@adityaambarik1513]

Can you provide solution for question 4?

[@bollywood316]

Using Redundancy Theorem HW Example 2: ans--> F = AB' + BC HW Example 4: ans--> G = (A+B).(B'+C)

[@book_worm4852]

thx bro for this amazing vid.

[@vinodeslavath2098]

Please show the proof of (A+B)(A+C)(B'+C)

[@SnidMist]

3rd point , does it mean only one variable out of the 3 should be complimented

[@pavanjyothireddy1048]

This may be old videos it's help me a lot

[@viratvihaan23]

6 years back only with out zoom u have gave wonderful lectures

[@rohitdeore1585]

Ex. F=AB'+BC+AC solution: F=AB'+BC EX. G=(A+B).(B'+C).(A+C) Solution: G=(A+B).(B'+C) thank you from bottom of my heart... Ur teaching is excellent one.... 🔥👏👍... I don't have any interest in this subject but now I am going to took interest in this subject....once again thank you so much sir.....

[@codelikealpha]

7 years old and still the best content

[@g.sabarimani1278]

thank you for showing this theorem sir,Thank you very much

[@adarshkumar3424]

can we use the redundancy theorem for the expression like this F=(A+B+C) (A+B'+C) (A+B+c') if so please explain it

[@harshiniadusumilli2509]

we can'y apply redundancy theorem becoz the condition number 2 is not satisfied

[@goshikasaravana]

@@harshiniadusumilli2509 agreed in the above example, the variable is repeated thrice

[@abu-bakarkhan7022]

Thnks a lot sir

[@naveenav8818]

for example no.2 - AB'+BC for example no.4- [A+B].[B'+C]

[@adenosinetp10]

H.W: F = AB'+BC+AC Ans: Using Redundancy theorem F = AB'+BC HW: G = (A+B).(B'+C).(A+C) Ans: Using Redundancy theorem G = (A+B).(B'+C)

[@soumdeeptaroy3341]

The teaching way is worth appreciating sir🙏

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Thank a lot sir u made computer easy for me Ur videos have great knowledge and content Thank you sir ...

[@Nafeel_Afri]

Thank you so much

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I am watching this video after 7 years but still is the best👍💯👍💯🎉

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thank you sir:) :) :) :) :) :) :) :) :) :) :) :) :) :) :)

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Thank you sir ! your explanation is way better than my teacher ! truly appreciated for your work ! It help alot alot alot !

[@HemanthKumar-pu1gi]

I am very thankful to you.you saved my exam. Keep posting for other subjects too.all the best. answer is---AB'=BC.

[@gopimanishankarnemmoju3990]

It's not AB'=BC ..it's AB'+BC .

[@ru_uwu]

hw 2 answer - ( B' + C ) . ( A + B )

[@srinivasulus3657]

Really ur teaching is awesome

[@HelloWorld40408]

Thank you Sir

[@kisitufahad9984]

G=(A+B).(B'+C)

[@priyanka.sarkar]

ii) AB' + BC iv) (A+B).(B'+C) Thanks for the video...

[@shubhamsatle8144]

Priyanka Sarkar How many boolean functions can be realized using n variables

[@zakeee1539]

dear sir your videos are really good.....i learn lot of things...sir kindly give some explanation on propagation delay...sir plz help

[@MohdAslam-yz5kw]

i am referring to the rule (iii) if more than one variable is Complemented, then what should we do? i am (not) asking about the case where all the variables are Complement

[@rohitnamballa236]

same doubt

[@cheapgamer4U]

can we break 3rd one further to A.C+B.A'

[@shahinalam5003]

Just Awesome Sir💗💗

[@sanjana9727]

Thank you sir😄

[@divagupta6611]

Thank u sir ! It is more helpful for me and future engenearings stydents ....🔥🔥❣️

[@MysteryJoy]

5:08 Ex 2 -- F =AB'+BC+AC Answer = AB'+ BC Ex 4 Answer = (A+B). (B+C)

[@KCEEE_VIPULKUMAR]

Example no 2 solution ----AB'+BC & Example no 4 solution ---(A+B).(B'+C)

[@anbuanbu2094]

Hi guys, any one tell me why we definitely add one in end of the bolean algebra for Y = A.B + A'C + BC This problem.

[@abdulbasithofficial4278]

Brilliant explanation

[@deepshikhagupta1977]

Thankyou sir, this video means a lot to students like me, like I had no interest in this topic electronic devices, not even my professor could help me but you. So thankyou very much for uploading this video and making this huge series...

[@adekoyajeremiah6145]

EX 4 ANS: G= (A+B) . (B' + C)

[@Manu-kj9gl]

hw answer ) 1.AB' + BC 2)(A+B).(B'+C)

[@John-oi9cn]

Great video.

[@mirknankazmzad8517]

Why BC is redundant in the first example? Can someone explain?

[@creative_new_1377]

Thanks a lot sir for these amazing videos