A community for students.
Here's the question you clicked on:
 0 viewing
rajat97
 one year ago
Mathematical logic
prove the following equivalence without using truth table:
~(p v q) v (~p ^ q) = ~p
rajat97
 one year ago
Mathematical logic prove the following equivalence without using truth table: ~(p v q) v (~p ^ q) = ~p

This Question is Closed

freckles
 one year ago
Best ResponseYou've already chosen the best response.3You can make a truth table.

rajat97
 one year ago
Best ResponseYou've already chosen the best response.0it is to be done without a truth table!

rajat97
 one year ago
Best ResponseYou've already chosen the best response.0i'm sorry i forgot to add it

freckles
 one year ago
Best ResponseYou've already chosen the best response.3for some reason I'm thinking demorgan's law

rajat97
 one year ago
Best ResponseYou've already chosen the best response.0yeah i did it and then tried to apply the distributive law but it was going way too long

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[(p \text{ or } q) \text{ or } ( p \text{ and } q ) \\ \text {recall } T \text{ or } Q \equiv (T \text{ and } Q) \\ \\ (p \text{ or } q ) \text{ or } (p \text{ or }  q) \\ \] going to try to use it again \[([p \text{ or } q ]\text{ and } [p \text{ or } q]) \] I guess you can try that distributive law too \[(p \text{ or } (q \text{ and } q ))\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3I think that should make it pretty easy on you now :)

freckles
 one year ago
Best ResponseYou've already chosen the best response.3you know since q and not q is definitely what kind of statement?

rajat97
 one year ago
Best ResponseYou've already chosen the best response.0that's how it is solved in the textbook but i don't understand how did you use the distributive law it should be −([p or q] and [p or −q]) is equivalent to [p and (p or q) or q and (p or q)] according to me

freckles
 one year ago
Best ResponseYou've already chosen the best response.3distributive law says: \[(P \text{ or } Q ) \text{ and } (P \text{ or } R) \equiv P \text{ or } ( Q \text{ and } R ) \\ \text{ where we have } \\ P=p \\ Q=q \\ R=q\] plug in

rajat97
 one year ago
Best ResponseYou've already chosen the best response.0i may be a bit irritating but i didn't get that

freckles
 one year ago
Best ResponseYou've already chosen the best response.3wait which part? do you understand that is one of the distributive laws?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3like and side the not we had: \[(p \text{ or } q ) \text{ and } ( p \text{ or } q)\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3this is in that exact form I mentioned above where P=p and Q=q and R=q

freckles
 one year ago
Best ResponseYou've already chosen the best response.3just plug into that law above it

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[p \text{ or } (q \text{ and } q)\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3and of course bring down the not sign that was in front of all of that

freckles
 one year ago
Best ResponseYou've already chosen the best response.3like inside the not we had*

Loser66
 one year ago
Best ResponseYou've already chosen the best response.2To me, it is quite simple not(p or q) AND (not p AND q = (not p AND not q) AND(notp AND q) Now, all of them are AND, we can take off the parentheses \(\neg p \wedge \neg q\wedge \neg p \wedge q\equiv \neg p \), since \(\neg q \wedge q =0\)

rajat97
 one year ago
Best ResponseYou've already chosen the best response.0no i didn't understand (P or Q) and (P or R)≡P or (Q and R) where we have P=pQ=qR=−q according to me it should be [P and (P or R)] or [Q and (P or R)]

rajat97
 one year ago
Best ResponseYou've already chosen the best response.0yeah i'm sorry i mean according to the thing i read in the textbook

freckles
 one year ago
Best ResponseYou've already chosen the best response.3oh I don't know about that rule that what I have to be something else I try to prove but I was using the distributive law which says https://en.wikipedia.org/wiki/Logical_equivalence Instead of typing it out again

freckles
 one year ago
Best ResponseYou've already chosen the best response.3also @Loser66 's thing looks good to me he made it a completely an and operation thing

freckles
 one year ago
Best ResponseYou've already chosen the best response.3oh that is because he changed the original statement

freckles
 one year ago
Best ResponseYou've already chosen the best response.3she sorry again @Loser66

rajat97
 one year ago
Best ResponseYou've already chosen the best response.0okay i'll try to get it just a moment

Loser66
 one year ago
Best ResponseYou've already chosen the best response.2hahaha. it's ok @freckles I worked on his/her original problem (on the very first post)

freckles
 one year ago
Best ResponseYou've already chosen the best response.3yeah the statement would have been easier if it was and in the middle

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0``` ~(p v q) v (~p ^ q) (~p ^ ~q) v (~p ^ q) ~p ^ (~q v q) ~p ^ 1 ~p ```

rajat97
 one year ago
Best ResponseYou've already chosen the best response.0thanks a lot everybody for your help we are allowed to choose only one best answer or else i would give each one of you a medal one last question please what if it is (p ^ q) v (q v p) how would the working with distributive law go?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.2=(p v (q v p)) ^ (q v (q v p)) = (p v q) ^ (q v p) = p v q

Loser66
 one year ago
Best ResponseYou've already chosen the best response.2=\(q\rightarrow \neg p\)
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.