[Discrete Mathematics] Logic Laws Examples

[@Trevtutor]

Check out my new course in Propositional Logic: trevtutor.com/p/master-discrete-mathematics-propositional-logic It comes with video lectures, text lectures, practice problems, solutions, and a practice final exam!

[@JuanDeSouza7]

4:49 "I don't know how we call this... whatever, I'll say it's a negation." The most sincere teacher ever! hahahah

[@محمد-م5ث1ش]

He really is funny

[@t0khyo]

@@محمد-م5ث1ش also you are a joke bro

[@t0khyo]

@@محمد-م5ث1ش are u studying at Tanta University?

[@esmaerlefrankie8515]

my lecturer really leaves it up to us to study for discrete structure so YOU ARE A GOD SENT THANK YOU MAN

[@marielkayeorlido1641]

Hello! I'm taking up Discrete Structures now, and I'm reaaaaally thankful for your videos! Your explanations are easy to understand! Thank you so much!

[@corporalwaffles]

At 2:14, I think it's supposed to be idempotent law

[@AlexThaBird]

It is - for people who comes by and read this. Use Idempotent here. It is notated: p ∧ p p. Thank you for pointing that out.

[@alijonemerchant2793]

@@AlexThaBird Thank you, that helped a lot! I would like to add that this law works both for AND and OR!

[@bestyoueverhad.2408]

i had first applied associative and then claimed redundance but thanks for reminding me of indempotence

[@nooneg4233]

At 2:15 theres a mistake ,indempodent law is used which states that :PVP==P,P^P==P so for P^(P^Q) ,we arrange it as( P^P)^Q as due to indempotent law P^P==P, so our conclusion is (P^Q) & next is so on with the video ..Btw thank you sir your videos are lifesaver🤍

[@louis9116]

7:38 the most elegant phi I've ever seen

[@dichocolate]

holy fuk

[@johnangeloperez9866]

so can we say that q -> q T is a definition by itself, like if A, then A is always TRUE?

[@sarkersaadahmed]

3:35 why isnt the first one idempotant law

[@aboutthereality179]

Thank You Trev for these examples.

[@darwinmanalo4934]

This is very helpful. Please do some examples (problems and solutions) for Venn Diagram. Thank you :)

[@edberaga6357]

I don't understand the distribution law with different connectives at 3:38 ... can you explain more...?

[@XeroKG]

he took the whole second bracket of (P and Q) and distributed it on each element of the first bracket

[@djswagmac7763]

At 5:12 can you just use the absorption law for step 4?

[@Trevtutor]

Yes. Some professors don't allow Absorption as a given law, though, so it's good to show it the long way.

[@raphaelgeronimo]

6:06 what even is "definition of the arrow"? I don't think I encountered it in your videos...

[@Snoopfrogg.]

Do you have any videos on Rules of Inference?

[@Carrymejane]

Before this video in the playlist

[@SO-oy2li]

One question. For the first example step 3, (p & q) v ((p & q) & not p) for the second bracket set, since it's pretty associative, can we really just remove the brackets inside and think of it as (p & q & not p) as well?

[@Trevtutor]

Yes, we can. I keep them in for illustrative purposes so it's clear why I can use the laws I do. Some instructors may also require you to keep those there, but you don't really need them.

[@nishantbisht4296]

@@Trevtutor how you write false after this line . i.e., where is q????

[@everchann]

what's the definition of the arrow?

[@shubhamgoswami3722]

~(p + q) + (~p × q) equivalent to ~p Prove it by laws + For or x for and ~ for negation

[@ES50678]

Is the law of the excluded middle the same as the inverse law? In a previous video where you defined the laws, you defined the inverse law as: P V -P = T, which at 7:00 you use but write it down as the law of the excluded middle.

[@kodymyler304]

do you have anything on Boolean Alg?

[@RobinHood-eu4er]

is there a way to figure out quickly which law i should use in every step ? because i always get stuck and cant simplify it seems very complicated !

[@Trevtutor]

It's really just practice and pattern recognition. Do DeMorgan's when you can, distributive when you can, and then check to see when others apply.

[@nightravels4028]

I struggled with this too. For example in this video's first question, when applying DeMorgan's law, it looked as if it could be applied to ~ (~ (p ^ q) rather than the whole thing, so you would have ~ ( ~ (p ^ q) v r ~ ( ~p v ~q) v r. Which in turn seems like it can be double negated to get (p v q) v r. But I drew the truth tables for both this answer and your correct answer and they are not logically equivalent, so clearly I used the laws incorrectly. Would you possibly be able to explain why this is an incorrect application of DeM's and how you should know how to use it like you did. Should it always be applied to the entire wff, rather than just part of it? Other than that, I can't express my thanks enough for these videos. They're absolutely fantastic, and such a huge help for my degree. Don't stop teaching!

[@C0URE]

Question; are we allowed to have not p v p, or is this rule strictly for not p and p?

[@Ackk567]

P v (~p ^ q ) can i use absorb law and the result is ~p . Is that true??

[@perpetualmamaba3750]

qustion, for the second example, couldnt you have used the Absorbtion law for the p^(p^q))?

[@perpetualmamaba3750]

nevermind, i just realized the signs have to differ

[@nooneg4233]

@@perpetualmamaba3750 bro🤍 ,indempodent law is used which states that :PVP==P,P^P==P so for P^(P^Q) ,we arrange it as( P^P)^Q as due to indempotent law P^P==P, so our conclusion is (P^Q) & next is so on with the video ..

[@pexma]

5:04 isn't it Inverse law? p ^ -p = F

[@abdifatahmoh]

Yes it's.

[@christophermalecki9772]

This video is super helpful. Question, can't you just say that (Q implies Q) is a T?

[@char_nette]

You still need an answer?

[@asilvap]

The result of the first example wouldnt make that expression ambiguos since there is no parentheses?

[@Trevtutor]

If the operators are all "and" or "or" then it's fine to omit the brackets due to associativity and commutativity.

[@asilvap]

Thanks!!

[@murigig]

I think your videos should be in the syllabus 😂😂 Just saying 😪✌️

[@ulysses_grant]

0:41 when you see emojis everywhere.

[@kennytran3816]

What are the hard brackets symbolizing in "not[(p and q) -> r]" ?

[@razeer1232]

Just another type of brackets functioning the same way as ( ... ). He used different brackets so people don't get confused.

[@rdzsystem1221]

thank you bro

[@MrKB_SSJ2]

1:37

[@MrKB_SSJ2]

0:00

[@ericgutierrez347]

I FUCKING appreciate you!

[@bekkiiboo619]

I'm having some trouble finding the intuition behind: p-->q = ~p v q.

[@Trevtutor]

You can solve this with a truth table or with the semantics of the conditional. p->q is true if p is false or q is true. That's the same as ~p v q.

[@bekkiiboo619]

I appreciate your reply so much! I understand I can use the truth table, but I want more than just the computational abilities... How does this sound: The conclusion is true if p is false, automatically. This is so, because with p being false, q can be false and the statement holds true still. OR Q can be true and the statement still holds true.

[@LiyosiCollins]

@@bekkiiboo619 I usually not try to get the intuition for p-->q = ~p v q because: - Once I get the intuition for p -> q statement, then, for its other logically equivalent statements, I always defer to using symbolic manipulations to arrive at them, without bothering to understand the intuition for them. This is so because symbolic manipulations sometimes results into some statements that are just hard to reason about

[@MwongelaJoseph-dt5ob]

would you consider not to skip steps please😂

[@e2k220]

thanks xqcL

[@MrKB_SSJ2]

5:42