## liliy 2 years ago discrete math:(attached)

1. liliy

2. liliy

am i doing it right?

3. liliy

@mathmate ????

4. satellite73

first of all the answer is definitely T

5. liliy

right... how?:(

6. satellite73

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 $$p\lor\lnot q$$

7. mathmate

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

8. liliy

right...

9. mathmate

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

10. liliy

oy.

11. mathmate

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

12. mathmate

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

13. satellite73

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

14. satellite73

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

15. liliy

so can you write it out?

16. mathmate

17. satellite73

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

18. liliy

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

19. satellite73

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

20. mathmate

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

21. mathmate

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

22. mathmate

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

23. mathmate

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

24. mathmate

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

25. mathmate

Would that be obvious now?

26. liliy

27. liliy

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

28. 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.

29. satellite73

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

30. satellite73

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 $$T$$

31. satellite73

it is hidden in the last line written by @mathmate

32. 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

33. satellite73

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 $$\lnot p \lor q$$

34. satellite73

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 $$p \lor \lnot p$$

35. liliy

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...

36. satellite73

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

37. liliy

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

38. satellite73

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

39. satellite73

well not really, i used a machine to do this

40. satellite73

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

41. satellite73

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

42. mathmate

Right you are!