Propositional Logic − Logical Equivalences

Discrete Mathematics: Propositional Logic − Logical Equivalences
Topics discussed:
1) Logical Equivalence definition and example.
2) Most common and famous logical equivalences.
3) Logical equivalences involving conditional statements.
4) Logical equivalences involving biconditional statements.

Music:
Axol x Alex Skrindo - You [NCS Release]

#DiscreteMathematicsByNeso #DiscreteMaths #PropositionalLogic

[jewelleaniag]

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

[dysphoricjoy]

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

[gudugudu]

I'm watching this for 7th time .. but still can't 💔

[vickypatel]

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🤣

[subhradipsaha]

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]

[PetitePhillyLife]

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

[jenweatherwax]

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

[OG_Truth_Teller]

The answer for home work problem in Biconditonal
Note: Symbols used ('^' and), ('√' or), ('->' conditional) and ('<->' Biconditonal)
Q] ~(p<->q) = p<->~q
Here is the solution
~(p<->q)= ~{(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

[yannmergezs]

P this p that imma drop this major brah

[AmandaSim-us]

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

[gobindaadhikari]

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

[Factsmotivation]

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!!!

[Liamlefe]

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

[Domenic]

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

[serra]

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

[anshikayadav]

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

[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

[bringhappiness]

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

[carlosdiaz]

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

[theophilus_pato]

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