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liliy Group TitleBest ResponseYou've already chosen the best response.0
am i doing it right?
 2 years ago

liliy Group TitleBest ResponseYou've already chosen the best response.0
@mathmate ????
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
first of all the answer is definitely T
 2 years ago

liliy Group TitleBest ResponseYou've already chosen the best response.0
right... how?:(
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
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 \(p\lor\lnot q\)
 2 years ago

mathmate Group TitleBest ResponseYou've already chosen the best response.2
Remember that de Morgan's law requires you to interchange the and / or, like: ~(~a or b) <=> a ^ ~b
 2 years ago

mathmate Group TitleBest ResponseYou've already chosen the best response.2
So there is a problem in the first term of the third line (second written line).
 2 years ago

mathmate Group TitleBest ResponseYou've already chosen the best response.2
The same problem with the second term. Otherwise, it will be the same as what you did.
 2 years ago

mathmate Group TitleBest ResponseYou've already chosen the best response.2
You skip a lot of steps, as @satellite73 pointed out. You may lose points in an exam.
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
\(\lnot (p\to q)\) \(\lnot (p\lor \lnot q)\) \(\lnot p \land q\) i believe
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
@mathmate probably has a better explanation, but as he/she said take it slow
 2 years ago

liliy Group TitleBest ResponseYou've already chosen the best response.0
so can you write it out?
 2 years ago

mathmate Group TitleBest ResponseYou've already chosen the best response.2
I will follow your thoughts.
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
\[\lnot[(p\to q) \land (q\to r)]\lor (q\to r)\] may be better, yes?
 2 years ago

liliy Group TitleBest ResponseYou've already chosen the best response.0
hold up. im gonna write it all out and you guys tell me if its right
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
but i kind of suck at this, so i will let mathmate check steps
 2 years ago

mathmate Group TitleBest ResponseYou've already chosen the best response.2
First replace all p>q etc with the or equivalents, ~[(~p or q) ^ (~q or r) ] or (~p or r)
 2 years ago

mathmate Group TitleBest ResponseYou've already chosen the best response.2
Now apply de Morgan's ~(~p or q) or ~(~q or r) or (~p or r)
 2 years ago

mathmate Group TitleBest ResponseYou've already chosen the best response.2
Apply de Morgan again: (p ^ ~q) or (q ^ ~r) or ~p or r
 2 years ago

mathmate Group TitleBest ResponseYou've already chosen the best response.2
Rearrange (commutativity of "or") ~p or (p^~q) or r or (q^~r)
 2 years ago

mathmate Group TitleBest ResponseYou've already chosen the best response.2
Now use associativity: [~p or (p^~q)] or [r or (q^~r)]
 2 years ago

mathmate Group TitleBest ResponseYou've already chosen the best response.2
Would that be obvious now?
 2 years ago

liliy Group TitleBest ResponseYou've already chosen the best response.0
BAsically im going to assume things to be True or false and then go on from there...
 2 years ago

mathmate Group TitleBest ResponseYou've already chosen the best response.2
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.
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
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 \(T\)
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
it is hidden in the last line written by @mathmate
 2 years ago

mathmate Group TitleBest ResponseYou've already chosen the best response.2
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
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
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 \(\lnot p \lor q\)
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
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 \(p \lor \lnot p\)
 2 years ago

liliy Group TitleBest ResponseYou've already chosen the best response.0
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...
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

liliy Group TitleBest ResponseYou've already chosen the best response.0
no, that's wat i did originally and then i realized that in the instructions my professor said specifically NOT to use tables :(
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
it would look like this \[\begin{array}{cccccccc} P & Q & R & P\Rightarrow{}Q & Q\Rightarrow{}R & (P\Rightarrow{}Q)\land{}(Q\Rightarrow{}R) & P\Rightarrow{}R & ((P\Rightarrow{}Q)\land{}(Q\Rightarrow{}R))\Rightarrow{}(P\Rightarrow{}R) \\ \hline 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 1 & 0 & 0 & 1 & 1 \\ 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ \hline 1 & 0 & 0 & 0 & 1 & 0 & 0 & 1 \\ 1 & 0 & 1 & 0 & 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 & 0 & 0 & 0 & 1 \\ 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ \hline \end{array}\]
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
well not really, i used a machine to do this
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

mathmate Group TitleBest ResponseYou've already chosen the best response.2
Right you are!
 2 years ago
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