discrete math:(attached)

- anonymous

discrete math:(attached)

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- anonymous

##### 1 Attachment

- anonymous

am i doing it right?

- anonymous

@mathmate ????

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- anonymous

first of all the answer is definitely T

- anonymous

right... how?:(

- anonymous

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 \(p\lor\lnot q\)

- mathmate

Remember that de Morgan's law requires you to interchange the and / or, like:
~(~a or b) <=> a ^ ~b

- anonymous

right...

- mathmate

So there is a problem in the first term of the third line (second written line).

- anonymous

oy.

- mathmate

The same problem with the second term.
Otherwise, it will be the same as what you did.

- mathmate

You skip a lot of steps, as @satellite73 pointed out.
You may lose points in an exam.

- anonymous

\(\lnot (p\to q)\)
\(\lnot (p\lor \lnot q)\)
\(\lnot p \land q\) i believe

- anonymous

@mathmate probably has a better explanation, but as he/she said take it slow

- anonymous

so can you write it out?

- mathmate

I will follow your thoughts.

- anonymous

\[\lnot[(p\to q) \land (q\to r)]\lor (q\to r)\] may be better, yes?

- anonymous

hold up. im gonna write it all out and you guys tell me if its right

- anonymous

but i kind of suck at this, so i will let mathmate check steps

- mathmate

First replace all p->q etc with the or equivalents,
~[(~p or q) ^ (~q or r) ] or (~p or r)

- mathmate

Now apply de Morgan's
~(~p or q) or ~(~q or r) or (~p or r)

- mathmate

Apply de Morgan again:
(p ^ ~q) or (q ^ ~r) or ~p or r

- mathmate

Rearrange (commutativity of "or")
~p or (p^~q) or r or (q^~r)

- mathmate

Now use associativity:
[~p or (p^~q)] or [r or (q^~r)]

- mathmate

Would that be obvious now?

- anonymous

##### 1 Attachment

- anonymous

BAsically im going to assume things to be True or false and then go on from there...

- mathmate

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.

- anonymous

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

- anonymous

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 \(T\)

- anonymous

it is hidden in the last line written by @mathmate

- 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

- anonymous

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 \(\lnot p \lor q\)

- anonymous

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 \(p \lor \lnot p\)

- anonymous

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...

- anonymous

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

- anonymous

no, that's wat i did originally and then i realized that in the instructions my professor said specifically NOT to use tables :(

- anonymous

it would look like this
\[\begin{array}{|c|c|c|c|c|c|c|c}
P
& Q
& R
& P\Rightarrow{}Q
& Q\Rightarrow{}R
& (P\Rightarrow{}Q)\land{}(Q\Rightarrow{}R)
& P\Rightarrow{}R
& ((P\Rightarrow{}Q)\land{}(Q\Rightarrow{}R))\Rightarrow{}(P\Rightarrow{}R) \\
\hline
0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 \\
0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 \\
0 & 1 & 0 & 1 & 0 & 0 & 1 & 1 \\
0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
\hline
1 & 0 & 0 & 0 & 1 & 0 & 0 & 1 \\
1 & 0 & 1 & 0 & 1 & 0 & 1 & 1 \\
1 & 1 & 0 & 1 & 0 & 0 & 0 & 1 \\
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
\hline
\end{array}\]

- anonymous

well not really, i used a machine to do this

- anonymous

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

- anonymous

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

- mathmate

Right you are!

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