[Discrete Mathematics] Logic Laws Examples 2

[@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!

[@kenmeyer100]

2. example: instead of introducing double negation in the 4th step, he could have easily used DeMorgan and could have been finished in the 5th step

[@pacman4521]

yes, not(q or p) = (not q and not p) We would go backwards in step 4

[@bluegiant13]

I did it as such: 1. (p -> r) /\ (q -> r) | Axiom | 2. (~p v r) /\ (~q v r) | Conditional 2* | 3. (r v ~p) /\ (r v ~q) | Commutativity 2* | 4. r v (~p /\ ~q) | Distributivity | 5. (~p /\ ~q) v r | Commutativity | 6. ~(p v q) v r | DeMorgan's | 7. p v q -> r | Conditional Is this legit? Did I make any mistakes?

[@thepedzed]

Looks fine, but you can get rid of step 3 + 4. Step 5 (as it is) would be the distributivity (and become renumbered to step 3).

[@ES50678]

So can we just "stick" a double negation where ever we want? like what was done in step 4 @2:55 ? I'm not clear on how this was done. I had used DeMorgans law before the definition of the arrow. So last three steps were: ~(qvp) v r, via DeMorgans. ~(~(qvp)) -> r, via the definition of the arrow, then I used the double negation law to end up with (qvp) -> r.

[@Trevtutor]

Depending on how your professor lets you play with the laws, you can stick a double negation anywhere and it doesn't change the truth value of a statement.

[@ES50678]

That's interesting. Thanks for the help! I'm on a strict software development track for my CS degree and this is one of the final courses I'm taking. I WISH I had taken it first, before any programming course. Your videos are the best, I've shared them with my class and I know they are helping everyone so much.

[@calamity560]

I think everyone got confused with the conditional law and the question. Firstly the question is asking for “(not p OR q)->r”, secondly the conditional law is (p->q) = (not p OR q) which can be written as (p OR q) = (not p -> q), you can prove these with truth tables. In step 4 the tutor is adding a double negation but you don’t need to, if you just follow the -> deff (or conditional law) then ((not p AND not q) OR r)) = not( not p AND not q) -> r, (In this case p = (not p AND not q), Demorgan’s law flips the OR to an AND then double negation removes the not. Hope this helps

[@javaexpertsa8947]

Guys the double negation is not necessary, it took me some minutes to understand this.You can pull in Step 4, the negation out of the braces and you end up with ~(q v p) v r , since this is equal to (~q 'and' ~p) v r, now ~(q v p) v r is nothing more then (q v p) --> r, last step use the commutative law and flip q and p, so (p v q)-->r. That's it. :)

[@bluegiant13]

I also end up doing it as you did.

[@phamvankhoa6117]

There is a way that solves the last example much more simple and faster. Just 4 steps However, thank you for your vids.

[@uzukiz3870]

can someone explain on how the first problem he goes from step 3 to step 4? im confused

[@kRystal69]

this rule: p → q = ( ¬p v q ) what he did was 3. ¬¬q v ¬p [ ¬(¬q v ¬p), this is same as, ¬(p → q) ] 4. ¬q → ¬p [ ¬(p → q) then he gave ¬ to both p and q ] I hope that helps.

[@ceromb]

I did mine like 1. (p→r) /\ (q→r) 2. (~p v r) /\ (~q v r) Conditional 3. r v (~p /\ ~q) Distributive 4. r v ~(p v q) DeMorgan's 5. ~(p v q) v r Commutative 6. (p v q) → r Conditional Did I do mine right?

[@edberaga6357]

when doing the distributive law, is the order matters? like can we do "(~p ^ ~q) v r" instead "r v (~p ^ ~q)" that later needs to add communiatative law..?

[@pacman4521]

im a bit late but yes you should be able to do that lol

[@aboutthereality179]

Thank You Trev.

[@nightravels4028]

As a couple of other comments have asked, what is the purpose of using the double negation in step 4? Is it just for shits and giggles?

[@Trevtutor]

p -> q is equivalent to ~p v q. Therefore, in order to get to ~p -> q, you must use the law on ~~p v q.

[@wirito]

And also for shits and giggles -.-

[@effy1219]

what is the introduction of "the definition of arrow"

[@Trevtutor]

p->q ~pvq

[@effy1219]

i knew this formula, i meant how does this come

[@effy1219]

TheTrevTutor thanks for the reply, i didn't expect a reply from host broadcast :)))

[@mistersir3185]

idk if i can be of any help but lemme try we have, p --> q we want to prove *p --> q ~q --> ~p* = ~p v q [acc. to Conditional Disjunction] = q v ~p [Commutative Law] =~q --> ~p (in other words ~q-->~p ~~q-->~p . But ~~q=q acc. to Double Negation) and we've proved that *p-->q ~q --> ~p .*

[@nightravels4028]

It's called the conditional law formally. At least I think it is. If I'm wrong please correct me.

[@bekkiiboo619]

I really appreciate you making these videos, they truly help tremendously. However, this first example did not make sense, unless you can distribute the negation through: ~(q -- > p) = ~q -- > ~p. Does this work?

[@Trevtutor]

No. ~(q->p) = ~p -> ~q.

[@aaronzewdu]

@@Trevtutor ?

[@isomsg7077]

thank you

[@francofazzolari7973]

(2.58) What if you go from: 1__ [.....] 4__(not Q and not P) or R (apply here morgan law backwards) 5__ not(Q or P) or R (and here apply the logical equivalence conditional) 6__(Q or P) --> R ??

[@MrKB_SSJ2]

1:26

[@MrKB_SSJ2]

0:00

[@zhenhui4386]

Not a good tutorial. He missed a lot of steps

[@djswagmac7763]

ZHENHUI I'm guessing you're the 1 dislike

[@wirito]

lol so true

[@TheSulaimanKhaled]

true. Can someone explain what he did between 3. and 4? 1:21

[@marvimilaqi6376]

@@TheSulaimanKhaled 4 years late but basically another idea is when u have q v ~p u can use commutative law again to have p v q which by using Conditional law gives us p --> q and by using Contrapositive law p --> q gives us ~q --> ~p because Contrapositive law says that p --> q = ~q --> ~p