Propositional Logic − Logical Equivalences

[@vickypatel6496]

What's more important than proving this thing is you gave a proper explanation of what a logical equivalent are🤦‍♂️ In my University the professor was just like read it this are formulas🤣

[@dysphoricjoy]

I have a test today and this confirmed I will be failing that exam.

[@JustNormalHito]

Same

[@drinetorshorts]

😂

[@Salamanca-joro]

Did u

[@drinetorshorts]

@@Salamanca-joro i got 23/30 DSGT IS EASY THOUGH🤣🤣

[@Salamanca-joro]

@@drinetorshorts I am taking this subject this semester and my first lesson was this week about this topic , idk this whole thing seems easy to me I hope it keeps being this way so i can get A 😂

[@jewelleaniag146]

you know, my mind is like yeah okay i get it but what again? hahahaha

[@abdulmateen7660]

fr

[@santhoshkumarr7446]

Hi

[@mariecriscaamud4130]

ifyyyyy🤐

[@maazfaridi5946]

Same

[@venusunbagcg6171]

@@maazfaridi5946 hey are u studying in country where speaking native english

[@jenweatherwax7113]

You are amazing! Thank you for making this so clear and easy to understand!

[@advaithkumar5966]

for absorption law (a), another way to think is that either p or p and q needs to be true for the expression to be true. Hence if p is false, the expression has to be false and if p is true the expression has to be true. So its equivalent to p. Similar logic holds for (b)

[@OG_Truth_Teller]

The answer for home work problem in Biconditonal Note: Symbols used ('^' and),('√' or),('->' conditional) and ('' Biconditonal) Q] ~(pq) = p~q Here is the solution ~(pq)= ~{(p->q) ^ (q->p)} Using demorgans law in RHS ~(p->q) √ ~(q->p) Using 5th stmt in conditional equivalence (p^~q) √ (q^~p) Let s=p , r=~q So, (s^r) √ (~s^~r) Using 3rd stmt in Biconditonal equivalence s r This is equal to "p~q" I think im correct Thank you

[@smammahdi]

Thanks man

[@TheUnKnown666]

@@smammahdi teko kaise pta 🧐

[@royalcanon7433]

Can you tell me how to prove that first biconditional statement

[@OG_Truth_Teller]

@@royalcanon7433 Hey buddy i think you missed this video on bi- conditional property in the playlist, here is the link of video below kzbin.info/www/bejne/m5mulWaBoq6Fidk Any way lemme help you out with simple example let us consider this example P: is a polygon with 4 equal sides Q: is a square So, if P is true and Q is true then the proposition is true If P is true and Q is false or P is false and Q is true then the proposition is false This seems a tricky one if P is false (not a polygon) and Q is false(not a square) This makes sense bcoz its not a square since its not a polygon I hope it helps ,Good luck 🎉

[@lynnewang8813]

Is (p^~q) = (~q^p) ?

[@GKNaidu-hb5zv]

now this is the kind of stuff which is complicated and simple at the same time due to this channel !!!!!

[@anjali-dasila]

Negation(p implies q) equivalence p and negation of q Using (p implies q) equivalence to negation of p or q Negation ( negation of p or q) equivalence p and negation of q Then use de Morgan rule P and negation of q is equivalence to p and negation of q

[@subhalaxmiaran3367]

Thanks yaaa

[@Domenic367]

I just want to say THANK YOU!! You`re doing a great job on your videos, keep up the good work!

[@PetitePhillyLife]

Better explanation then the one my professor gave or what's in the trash book they made us buy

[@Mujahed.]

Same situation bro

[@hetaeramancer]

how much bro

[@hariszaib2728]

can anyone explain that in proving the absorption law, when he took p as common why did he change the signs.. like ^ to or and or to ^ .. time 5:36

[@kurmasaradhi8724]

​@@hariszaib2728 As mentioned here p^1 =p, so the 4th step can be written as (p^1)\/(p^q). Then by taking p as common by distributive law, we get p^(1 \/q)

[@002dd]

Loll, sammee

[@enasgeravi4372]

شكرا و جزاك الله خيرا

[@serra7379]

Your voice is like Rajesh Koothrappali’s. Thank you for video

[@theophilus_pato]

Wow have had a great understanding of everything.Thanks for the good work.

[@yannmergezs2820]

P this p that imma drop this major brah

[@bringhappiness3862]

Really helpful and yes I have done my home work ☺️😁 Thank you sir 😊

[@F_F9F]

احسنت الشرح جدا بسيط وواضح ❤️

[@ayaayucha5310]

🥰🥰🥰

[@QasimNizam-mk5px]

❤

[@snotface8]

Great video! I loved how you went in depth and proofed all logical equivalences!

[@gobindaadhikari3319]

Sir, please post the solution to homework problems in the description box so that we can verify or modify our solution

[@abe22er]

what is the link of the homework?

[@lakshmi1135]

@@abe22er it is given the end of video

[@Liamlefe]

In last question ~(pq)=p~q proved and it is also = ~pq Just check anybody please

[@PositivePulse288]

Amazing way of teaching...

[@helptech1642]

Your teaching is so nice or understanding thank

[@bozeiky]

For the homework #5 I got, -(p -> q) == -(-p v q) {Conditional Law} == - -p^-q {DeMorgans Law} == p^-q {Double Negation Law} Please let me know if this is right or how I did!

[@rgbkedits137]

This is correct

[@thirstymosha1147]

you are so much better than my lecturer thanks for existing

[@miriamDev]

Thank you Neso Academy, detailed explanation, but I have a very confusing question and quiet difficult to break it down, don't know if you can help out

[@yasserfathelbab1534]

lol so you're just whining or what

[@yasserfathelbab1534]

that's not how math works

[@raannnggggaaaaa]

@@yasserfathelbab1534 chill out man

[@yasserfathelbab1534]

@@raannnggggaaaaa Two years now and not a soul knows what the question is.....

[@muhibali205]

@@yasserfathelbab1534 still now I didn't get answers of that questions from homework.

[@jennysanchez822]

For the homework #4 involving bi-conditional statements like the following: "-(p q) = p -q " For this one I broke it like this "(-p -> -q) or (-q -> -p) = (p-> -q) (-q-> p) . Based on the Double negate law, " (-p-> -q) or (-q-> -p) is True as well as (p->-q) (-q-> p) which are True because no matter what if p is false it doesn't matter if q is False or True, p->q will always be TRUE. This is why they are equivalent. Any one want to give suggestions if I'm the right track here?

[@jerrinjose9633]

i think you r r18

[@hariszaib2728]

can anyone explain that in proving the absorption law, when he took p as common why did he change the signs.. like ^ to or and or to ^ .. time 5:36

[@shadow3491]

@@hariszaib2728 it wasn't p as common it was p^p=p

[@AidahBlessed-z7m]

How can I master these laws and how hey are applied

[@rezmir7115]

Very simple yet effective explanation. Thanks a lot!

[@subhradipsaha9518]

Last question. NOT (p biconditional q) is equivalent to (p biconditional NOT q) NOT (p biconditional q) = NOT ((NOT p AND NOT q ) OR (p AND q)) [Biconditional into implication into combination of NOT, AND, OR] = (NOT p OR NOT q) AND (p OR q) [DeMorgan's Law] = (NOT p OR NOT q) AND (NOT(NOT q) OR p) [Double Negation Law] = (p implies NOT q) AND ( NOT q implies p) [Implication] = p biconditional NOT q [Biconditional]

[@offlinemoe]

Hey bro Can you text me on instagram pls This is my acc ha_a_21.11

[@arichullai5626]

can you explain this line = NOT ((NOT p AND NOT q ) OR (p AND q)) [Biconditional into implication into combination of NOT, AND, OR]

[@subhradipsaha9518]

@@arichullai5626 A B is logically equivalent to (A -> B) AND (B -> A) which is logically equivalent to (NOT A AND NOT B) OR ( A AND B)

[@arichullai5626]

@@subhradipsaha9518 thank you for that but A-> B is logically equivalent to NOT A OR B.........

[@muhammadhilwan7406]

from where u got the (NOT(NOT q) OR p)?

[@mjlplofs4hpn253]

The solution of first h.w. is :¬(p_>q). Because we took ¬ out and replaced ^ with _>, So we equivalent the RHS and the LHS.

[@BinshadRayhan]

very simple explain

[@dimabraginskiy2969]

pls check if i am correct with 5th homework task: NOT(p=>q) p and NOTq We can transform it as following: (we can do a double negation of both sides) NOT(NOT(p=>q)) NOT(p and NOTq) (p => q) (NOTp or q) (right side is the same as 1st conditional statement) please correct me!

[@realhumanoid1323]

for the 5th : we have ¬( p -> q ) = p ^ ¬ q ; ................1 we know from the 1st proof that : ( p -> q ) = ¬ p ν q , therefore substituting this same value to : [ ¬( p -> q ) ] we get : ¬( ¬ p v q ) = p ^ ¬ q ; ..............................2 Now by DeMorgan's Law : ¬( p v q ) = ¬ p ^ ¬ q by applying demorgan's law and solving the 2nd equation we'll get : ( p ^ ¬ q ) = ( p ^ ¬ q ) hence therefore, LHS = RHS

[@Oogwood]

This is the hardest part of logic

[@DRASERUS]

Would be nice if you put the correct answer to your question in description so i can confirm if my answer is correct or not.

[@Factsmotivation2002]

10:24 Not(p->q) =p^not(q) =not(not(q) -> not(p)) =not(p->q) where p=not(q) and q=not(p) Therefore, Hence Proved!!!

[@jennysanchez822]

For the Homework (5): is -(p->q) = p and -q are equivalent because if you break it down like following:" -p implies - q" makes that statement True while "p and -q" makes the statement TRUE. Because they both have true values makes the statement true and equivalent. Is that why? Can someone explain to me or check if I'm the right track? Thank you in advance!

[@debjyotiray8364]

You have applied theoretical knowledge of the understanding of the equivalence, I guess it's correct!! The more simpler way that I used is to use De Morgan's laws that sir initially explained to prove it and it becomes just a three liner proof! Hoping that helped... Welcome in advance

[@Gupatik]

now you've said that "-p" implies "-q", and that's true so from there we can agree that "q" implies "p" which is also correct. but the problem is that we can't return back and say that "-q" and "p" are equivalent, the equivalent here is p and "q'' . the homework itself for me is not logical WHY, coze we have -(pq) which is also (pq) but not (p-q). let's make things even more simple, we have "q" and "p", both are equivalent then we say that "p" and "-q" are also equivalent which make no sense like if "a" is "a" then we say "a" is "-a which stands here for another alphabet different from a" and that's not so true.

[@shubikshashubi4971]

Thank you so much for explaing the laws which is more helpful in solving the problem thanks a lot

[@azharosaaf2034]

Thank you so much 💕💕💕💕💕💕💕 I am from India and I enjoy to see your vedios.

[@MathPlusAi]

Thanks for your best tutorials!!🤩

[@dagim6625]

great video!

[@Justinlabry]

I solved the HW and understood why there is no posting of it. haha :D To give you a hint, it is pretty lengthy. yo. My humble respects to the teacher.

[@SazzadRashid]

Can u plz give me the solution

[@mochi464]

omg its so complicated

[@ayaayucha5310]

Thank you so much ❣❣💯💯❣❣ I am from Algeria and I enjoy to see your vedios🥰🥰🥰

[@pubggaming5889]

Hii, wr is Algeria..

[@pavanyendluri]

Thank you Neso Academy

[@fatimalmasri5943]

Dude i owe u !! ❤ thanks Subscribed !

[@thirumeniparthiban6261]

Excellent !! Excellent!! nothing else to say. You please do some video lectures ( even paid ) on model checking buchi automata etc.,

[@СанжарАлманов-т3с]

love your videos man, really helpful, thank you veru much!!!

[@TowerBooks3192]

You just saved me. Thank you for this video

[@AmandaSim-u7s]

At 5:36 I'm confused as to what it means "taking p as a common". I see that p is converted to 1 and I'm confused as to how that happened

[@MrVrtex]

Exactly my thought. The 4th step and the question is exactly same. How does this work?

[@arhamkhxn]

I also have this confusion @nesoacademy please help!

[@norzuraika3892]

Thanks sir.. Really helpfull 👍 👍

[@AidahBlessed-z7m]

Sir, prove for me -(PvQ)^-p)=>Q

[@enasgeravi4372]

Thanks

[@juskonajak706]

I don't quite get how you solve the equivalences of biconditionals differently. The solution to number two is different from number three. In number two, you negated both p and q, but in number three you negated only p for the same conditions.

[@jamesmccloud7535]

Thank you! This helped me a lot!

[@VenusRahaf]

Thank you so much

[@studentperfect8135]

Thank you sir

[@mprl819]

Thanks, this cleared things up for me

[@venusunbagcg6171]

Hey which level(9/10/11grand) subject is this

[@nitishgautam5728]

5:37 😂 4th line is same place where you started to prove formula . By the way I like Boolean notation more , it's easier because we are familiar with + , .

[@houssam5180]

Better than in college. Thank you.

[@victoriarumbidzaimhlanga4934]

Thank you....

[@sirichandana605]

Explanation is well done 👍 sir

[@vaishnavimendre469]

Thank you sir🙌🏻

[@anshikayadav7857]

Sir can you please answer the explanation of question 5th (homework) in biconditionals.??

[@skycirnsnewaccount9225]

Thank you Indian Guy(i dunno what's your race is) but it really help me a lot since I have midterm exam today

[@sherazakbar6544]

Are we prove them with the help of truth table??

[@1832naipa]

Thanks sir alot😀

[@carlosdiaz9998]

5:38 What does "Taking P as common" mean?

[@vandanapathak8453]

Do you know now?explain me

[@jamunajai2165]

Thank you very much sir.... clearly understood...... Excellent explanation.....

[@mimischly2547]

Awesome man

[@tonoyislam718]

Thank u so much for help me

[@Momin_73]

I get all of this but do i have to memorize these laws

[@esuendalewdebebe7991]

can you solve to me (p^q)/bi implies p and p=>q/implies in logical equivalence if p,q and r use a truth table please ?

[@BCS__NimraHashmi]

Outstanding

[@JudeGussman]

Thank you!

[@harishdasari4808]

great explanation

[@Hahahahahahahhahahahhahahshsha]

If only the test is this easy

[@cautionseaman]

“Right”

[@humerashaikh8402]

Thank you sir all of my doubts finally got cleared

[@anilover5159]

What laws should I use when I now have 3 propositions? for example... ~(p^q^~r) ^~(~p^q^~r) ^~(~p^~q^~r)

[@deemaha8645]

LIFE SAVER!!!!

[@Ududushshdhssj]

is there any sites where we can practice these questions

[@spug03]

Thank you so much my dude

[@venusunbagcg6171]

In which class is this subject covered?

[@kunalgoher5008]

@@venusunbagcg6171 betch cse in sem 4

[@khansaparween7209]

Superbbbb sir...

[@nicholasstamatakis]

Awesome explanation!

[@Somerandomnessvvv]

Made me realize how simple it is.

[@dhananjaysangle7982]

HELPFUL VDO

[@viveksingh7388]

Sir please upload signals & systems lecture

[@zulaikhazainuddin2773]

Before this im still confusing about law of logical equivalences. Watching this before exam

[@kleur3356]

Can I ask if a distributive can be two statements only, something like (p v q) ^ ¬p ≡ ¬((p v q) →p)?

[@hosamessa9069]

Yoo thanks 👏

[@TitanTubs]

9:45 my dummy self would have put a double negation on the p. But I guess it is commutive(you can switch the p's and q's) Would it be wrong to do that? Is it only logically equivalent if you give it no one that isn't already negated?

[@dresscheme2940]

where did he get the ~(~q v ~p)?

[@Uaiaoahq]

So the three vertical lines is "="

[@arshidbhat7358]

Well That was Cool

[@anythingbeyondlimit8398]

thank you very much dude, I am just learning this for myself as I have graduated from school long time ago. but bruh! wat in the multiverse is this shoot! Aliens laugh at us with all these nonsensical convention we brought. I am learning and laughing at this!

[@soniaverma7870]

Tysm:)

[@trollface1994]

DawDology !

[@nomannoman2492]

Super

[@slater-cguy]

5. ~(p -> q) = p ^ ~q ~(~p v q) p ^ ~ q De Morgan

[@story_time_bd2512]

Can u tell what will be (not p V T) Is it true? what will be the ans of (not p V not q V q)

[@omerabbas2346]

please teach me!

[@prudhviraj4066]

Exellent bro