Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

discrete math:(attached)

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SIGN UP FOR FREE
1 Attachment
am i doing it right?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

first of all the answer is definitely T
right... how?:(
looks like you are on the right track, it is hard for me to read it all you are turning each implication of the form \(p\to q\) into a statement of the form \(p\lor\lnot q\)
Remember that de Morgan's law requires you to interchange the and / or, like: ~(~a or b) <=> a ^ ~b
right...
So there is a problem in the first term of the third line (second written line).
oy.
The same problem with the second term. Otherwise, it will be the same as what you did.
You skip a lot of steps, as @satellite73 pointed out. You may lose points in an exam.
\(\lnot (p\to q)\) \(\lnot (p\lor \lnot q)\) \(\lnot p \land q\) i believe
@mathmate probably has a better explanation, but as he/she said take it slow
so can you write it out?
I will follow your thoughts.
\[\lnot[(p\to q) \land (q\to r)]\lor (q\to r)\] may be better, yes?
hold up. im gonna write it all out and you guys tell me if its right
but i kind of suck at this, so i will let mathmate check steps
First replace all p->q etc with the or equivalents, ~[(~p or q) ^ (~q or r) ] or (~p or r)
Now apply de Morgan's ~(~p or q) or ~(~q or r) or (~p or r)
Apply de Morgan again: (p ^ ~q) or (q ^ ~r) or ~p or r
Rearrange (commutativity of "or") ~p or (p^~q) or r or (q^~r)
Now use associativity: [~p or (p^~q)] or [r or (q^~r)]
Would that be obvious now?
1 Attachment
BAsically im going to assume things to be True or false and then go on from there...
Your last approach is called proof by cases. Since there are three variables, you will need to have 8 cases, which is equivalent to a truth table. The idea here is to do a proof without a truth table.
allow me to butt in a second if you were going to prove this using a truth table, all the above work would be unnecessary you would simply construct the truth table for this (it might take a while) and observe that you get all T in the last column
your job doing this method is to try to come up with an expression that looks like \(p\lor \lnot p\) which you can then replace by \(T\)
it is hidden in the last line written by @mathmate
Let's take it from: [~p or (p^~q)] or [r or (q^~r)] use distributivity: [(~p or p) ^ (~p or ~q)] or [(r or q) ^ (r or ~r)] which reduces to [T ^ (~p or ~q)] or [(r or q) ^ T] or (~p or ~q) or (r or q) => ~p or r or (q or ~q) => ~p or r or T => T
like that. hope the method is clear. if you wanted to work from cases there would be no reason to change for example \(p\to q\) into \(\lnot p \lor q\)
when you do these without truth tables and you want to prove something is true, (at the risk or repeating myself) try to look for \(p \lor \lnot p\)
woah, this is intense... reading what you guys did, i undertsand it all.. but is there a way to "learn this stuff" bec my teacher did NOT do this...
if this was not the method, then as i said you could write out the entire truth table and see that the last column was all T
no, that's wat i did originally and then i realized that in the instructions my professor said specifically NOT to use tables :(
it would look like this \[\begin{array}{|c|c|c|c|c|c|c|c} P & Q & R & P\Rightarrow{}Q & Q\Rightarrow{}R & (P\Rightarrow{}Q)\land{}(Q\Rightarrow{}R) & P\Rightarrow{}R & ((P\Rightarrow{}Q)\land{}(Q\Rightarrow{}R))\Rightarrow{}(P\Rightarrow{}R) \\ \hline 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 1 & 0 & 0 & 1 & 1 \\ 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ \hline 1 & 0 & 0 & 0 & 1 & 0 & 0 & 1 \\ 1 & 0 & 1 & 0 & 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 & 0 & 0 & 0 & 1 \\ 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ \hline \end{array}\]
well not really, i used a machine to do this
don't be put off by all this nonsense. the statement itself is completely obvious. like saying if r is contained in q and q is contained in p then r is contained in p doh
you just have to grind it til you find it, there is no other way. write out what you know and see what you can find. practice will help, but of course in real life no one ever does such a thing so it will only serve you for your next exam. then you can forget about it. until the final
Right you are!

Not the answer you are looking for?

Search for more explanations.

Ask your own question