Boolean Algebra 3 - De Morgan’s Theorem

This video follows on from the one about simplifying complex Boolean expressions using the laws of Boolean algebra. In particular this video covers De Morgan’s theorem and how it can be applied, along with the other laws, to simplify complex Boolean expressions. It includes worked examples and some exercises you can try yourself, along with solutions.

[@dickstephenson]

So professionally done. This is what KZbin was made for.

[@ComputerScienceLessons]

Thanks for the comment. Much appreciated :)KD

[@joshuawaterson8268]

Excellent series, what a great help it's been so far! Thanks for publishing them publicly.

[@mraristocrat1074]

This channel deserves a million subs. The videos are very to the point and helpful. I've learnt a lot from these.

[@Szoki666]

God bless this video series, it helps me to pass Computer Science class. Thanks for the info sharing, have a super day!

[@Szoki666]

Really! At first, I did not even understand the whole logic. With these videos I can practise a lot, and I am so happy that I cannot stop it :)

[@ComputerScienceLessons]

It's great to hear they have been so useful. :)

[@sanaalshaar5406]

Thank you. I am enjoying myself greatly working through these videos.

[@jeremymcdonald5281]

I am in a 5week accelerated program for Mechatronics Engineering. Part of Mechatronics is computer science. I am in a Digital Systems 1 class. This is about as hard as it has gotten so far, and I am a junior. Drinking through a waterfall is more like it than a firehose as Iwould normally say. I am definately thankful for these videos you have made. I would be 100% lost without them. Everything is so well explained and put together and clean and concise. My text book for this course is just hard to look at in general. Thank you for these videos. I will definately be revisting them time and time again, I will not be able to master this subject in 5 weeks the way I would like, but I am already scribbling these videos into my future schedule until I can pound these expressions down with quickness and confidence. Thnak you again Computer Science! You are awesome!

[@ComputerScienceLessons]

That's music to my ears. You are most welcome :)KD

[@katekuhl6115]

Many thanks for the video, excellent resource for use with A-level CS students

[@ComputerScienceLessons]

You are most welcome. It's nice to hear from another A level CS teacher. If there's anything else you would like me to cover, please let me know :)KD

[@DrBlyat]

Thank you so much for all these videos man! Definitely valuable resources for those who didn't really pay attention during class, lol.

[@user-up8ut9cr2h]

This video need more likes for his work.

[@therealkon_]

FINALLY some who explains each step of the solution! I couldn't find any video or tutorial about this in my first language. Each time the teacher jumped through the steps like it was all self explanatory, so eventually i started to think i was too just dumb to understand. With your videos i finally get it!! Thank you for giving me hope again ;) , especially by saying it takes time to develope this skill!!

[@ComputerScienceLessons]

You are very welcome. Stick with it :)KD

[@avanith3579]

Thank you very much for these videos, you taught Boolean Algebra very well.

[@ComputerScienceLessons]

You are very welcome :)KD

[@Afatsom]

YOU ARE AWESOMEEE !!! MUCH LOVE

[@ThemisTheotokatos]

great lecture, thank you!

[@yawsokpor3061]

This is very important and one must understand

[@1shakezula1]

Thank you, this series was very helpful.

[@icebeardoesnttalkmuch8919]

that was a good video along with practice set. i have a question: The old DLD has been replaced by CMOS technology so are these laws (DeMorgans) still used in CMOS or are they obsolete?

[@swx6074]

Wow, the execution of the series is just perfect

[@ComputerScienceLessons]

Thank you :)KD

[@mostafaaldgeir4538]

thank you a lot.I spend a whole day looking for a good explamention

[@ComputerScienceLessons]

You are most welcome :)KD

[@johnzachlim]

Kaway-kaway sa mga gikan sa Moodle. ✋✋✋

[@Mel-jp5vb]

Thank you for the amazing video!

[@ComputerScienceLessons]

You are very welcome. :)KD

[@JazzyArtKL]

Hi Kevin, how would you apply De Morgan to simplify this one ~a~b+~bc+ac?

[@flandman8347]

At 11:48 you could remove the brackets which would result in (X V Y) ^ `X ^ Y. Using the absorptive law (X V Y) ^ Y = Y. So `X ^ Y

[@hugofriberg3445]

I find: ¬(A ∧ B) = ¬a ∨ ¬b really similar to: -(a + b) = -a - b Since both expressions switch the middle operator, and negates the values of A and B.

[@ComputerScienceLessons]

Good observation 💡 Perhaps Augustus De Morgan was thinking along the same lines :)KD

[@user-mk9tq9mj8c]

thank you so much

[@ComputerScienceLessons]

You're very welcome :)KD

[@LostInLeiden]

Do you have anything on consensus theorem? I'm trying to understand why X' + XY = X' + Y

[@repenttoreflect8933]

Your are excellent Mr professor

[@ComputerScienceLessons]

Thank you. :) KD

[@repenttoreflect8933]

How many inputs are required for a 6-output decoder? How can one write the logical equations and its corresponding circuit? Help out please Sir

[@budalasreenidhi8242]

Thank you so much

[@ComputerScienceLessons]

You're welcome :)KD

[@christianperez8414]

🐐🐐🐐 channel

[@ComputerScienceLessons]

Thank you (I think) :)KD

[@darrentravi6069]

On the second expression at around 6 minutes into the video could you have gone the other way and converted the AND to OR's?

[@ComputerScienceLessons]

More than likely - there's more than one way to crack a nut. In my videos I am trying to illustrate different techniques, rather than the shortest route to a solution (well, that's my excuse anyway!) :)KD

[@luxfortis4377]

Thank you for the amazing lecture. I think solution 3 can be reduced more to A+~B*~C

[@ComputerScienceLessons]

Quite possibly. Thanks for the comment. :)KD

[@luigigoterelli7888]

Great video by the way but l think the answer to the last question is not right if l am not mistaking. In 5th row you have x + y~xy therefore why didn't you erase that y because and operation is more important and you can reduce number of y's and you will end up with x + y~x, and then (x+~x)(x+y) = x+y. Please correct me if l am wrong. Sincere regards sir for the professional video. Everything else was perfect.

[@jaypark6812]

It’s a good channel, I just don’t understand why few people subscribe it.

[@tusharmahajan6833]

At 2:54 how can you just swap the OR for the AND? Wouldn't that change the meaning of the expression?

[@ComputerScienceLessons]

It's OK to do that as long as you do something else as well. Take a look at my video on logic gate combinations, there is a proof of de-morgan's in there. kzbin.info/www/bejne/in3PZXhpfrtpkJI

[@codythompson9973]

@@ComputerScienceLessons The link leads to a private video that we cannot view now.

[@ComputerScienceLessons]

@@codythompson9973 I do apologise. Please search my channel for Logic Gate Combinations

[@LMinett]

Was this re-uploaded?

[@ComputerScienceLessons]

Yes. There was a silly error explaining the algorithm which someone kindly pointed out. Fixed now.

[@LMinett]

Excellent. Please do keep up the good work. My pupils especially appreciate these videos during their revision.

[@ComputerScienceLessons]

It's great to hear my videos are helping other people as well as my own students.

[@bensonkwok951]

Help can't seem to solve this problem: x*z+(notx)*y+yz The answer is x*z+(notx) Note that * means and; + means or

[@ReZhorw]

That seems to be the consensus theorem, search for it on youtube and you'll find good explanations. Practically it means that you can remove the" + yz " term from xz + x'y I believe the answer should be xz + x'y and not xz +x' as you've written. Swapping places with y and z would be preferred Try multiplying the yz term in xy + x'z + yz with 1 and write the 1 as (x + x')

[@mukeshkrmahto8500]

Kmap ka video banayiye sir

[@ComputerScienceLessons]

kiya hua :)KD kzbin.info/www/bejne/aafOfpqKos6YkKs