Here's the question you clicked on:
liliy
discrete math:(attached)
first of all the answer is definitely T
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
So there is a problem in the first term of the third line (second written line).
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
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?
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