Question

discrete math:(attached)

Answer 1

am i doing it right?

Answer 2

@mathmate ????

Answer 3

first of all the answer is definitely T

Answer 4

right... how?:(

Answer 5

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 \(p\lor\lnot q\)

Answer 6

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

Answer 7

right...

Answer 8

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

Answer 9

oy.

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

so can you write it out?

Answer 15

I will follow your thoughts.

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

Would that be obvious now?

Answer 25

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

Answer 26

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.

Answer 27

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

Answer 28

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 \(T\)

Answer 29

it is hidden in the last line written by @mathmate

Answer 30

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

Answer 31

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 \(\lnot p \lor q\)

Answer 32

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 \(p \lor \lnot p\)

Answer 33

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...

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

well not really, i used a machine to do this

Answer 38

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

Answer 39

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

Answer 40

Right you are!