## Esmy 3 years ago Use truth tables to test the validity of the argument. p → ~q q → ~p ∴ p ∨ q

1. Esmy

this is valid

2. Esmy

Case p=T and q=T [(p -> ~q) ^ (q -> ~p)] -> (p V q) [(T -> ~T) ^ (T -> ~T)] -> (T V T) [(T -> F ) ^ (T -> F)] -> T [ F ^ F ] -> T F -> T T

3. Esmy

All four cases come out T, so the argument is valid. I think lol.

4. A.Avinash_Goutham

u need to verify that using a truth table?

5. Esmy

p q ~p ~q p -> ~q q -> ~p p v q T T F F F F T T F F T T T T F T T F T T T F F T T T T F Hmmm... I change my answer its invalid

6. A.Avinash_Goutham

then u need to make a dumbo table with all the combinations of p and q and verify it or u can use this p->q = (~p) v q

7. Esmy

"p -> ~q" and "q -> ~p" the last row has T and T so the conclusion is false?

8. A.Avinash_Goutham

actuall it mean that u shud conside the cases on when both of them are true

9. Esmy

Oh I totally miss understood the question

10. Esmy

So how would you do that? :o

11. Esmy

help can you explain to how you do it :/

12. A.Avinash_Goutham

consider those case when p → ~q q → ~p are true find them

13. Esmy

can you explain like in steps :/ I understand in steps.

14. A.Avinash_Goutham

hmm consider those inputs for which p → ~q is true q → ~p is true

15. Esmy

umhmm :o

16. A.Avinash_Goutham

lol sorry....... u shud try and understand

17. A.Avinash_Goutham

ok fine q → ~p is true for first three columns in the table

18. Esmy

uhmm :o

19. Esmy

p q ~p ~q ∼p∨q (∼p∨q)→ ∼q ———————————————————————————————— T T F F T F T F F T F T F T T F T F F F T T T T Like this :D

20. Esmy

Help D:

21. A.Avinash_Goutham

u dont need help gal jus think

22. Esmy

Blagh I am thats why I'm asking if I did right or not

23. Esmy

Okay I'm guessing i didn't do that one right so ill work on another one.

24. A.Avinash_Goutham

lol draw truth tables for p → ~q q → ~p and pvq see when p → ~q q → ~p are true pvq is true

25. Esmy

(~p V q) --> ~q (~T V F) --> ~F (F V F) --> T F --> T T So (~p V q) --> ~q is true.

26. Esmy

there

27. Esmy

Blagh W/e I'm right I've done everything... :l

28. Esmy

q p (q → ~p) → (q ∧ ~p) T T T T F T F T F F F F

29. satellite73

the first two statements are identical $p\to \lnot q\iff q\to \lnot p$ one is the contrapositive of the other

30. Esmy

hmmm is my table okay :/ ?

31. Esmy

I've done like three so far -_-

32. satellite73

let me right it correctly $\begin{array}{|c|c|c|c|c|c} P & Q & \lnot{}P & \lnot{}Q & P\to\lnot{}Q & Q\to\lnot{}P \\ \hline T & T & F & F & F & F \\ T& F & F & T & T & T \\ F & T & T & F & T & T \\ F & F & T & T & T & T \\ \hline \end{array}$

33. satellite73

write

34. phi

the final step is put in the column p v q (this is the first problem) An argument is INVALID if and only if it is logically possible for the conclusion to be false even though every premise is assumed to be true.

35. satellite73

if you want to check your truth table, use this nice site here takes a second to get used to 0 and 1 instead of T and F, and a minute to learn the syntax but it is really useful http://www.kwi.dk/projects/php/truthtable/?

36. satellite73

but try to think "logically" meaning like a human being instead of using T and F the first statement and the second statement are identical, one is the contrapositive of the other. so you do not need them both, they say the same thing

37. Esmy

so none of my tables where right :/ ...

38. phi

this looks ok, The argument is invalid because of the last row p q ~p ~q p -> ~q q -> ~p p v q T T F F F F T T F F T T T T F T T F T T T F F T T T T F Hmmm... I change my answer its invalid

39. satellite73

P: it is raining Q: i will go to the store $$P\to \lnot Q$$ if it is raining then i will not go the the store $$Q\to \lnot P$$ if i go to the store, then it is not raining they are identical statements from that, can we conclude that is is raining or i go to the store?

40. satellite73

of course not! maybe it is sunny and i stay home anyway!

41. Esmy

Oh okay :o... I c I c

42. phi

I don't see the other problems..... with premises and conclusion

43. Esmy

neither do I

44. phi

I mean, did you post them?

45. Esmy

yeah I did ?

46. phi

Is this one of the problems (~p V q) --> ~q ? with premise ~p v q and conclusion ~q

47. Esmy

I posted my answer and everything D:?

48. phi

because it looks invalid p q (~p V q) ~q 0 0 1 1 OK 0 1 1 0 INVALID 1 0 0 1 OK 1 1 1 0 INVALID

49. Esmy

did you do this one or I did I forgot...

50. phi

somewhere way up above you typed (~p V q) --> ~q (~T V F) --> ~F (F V F) --> T F --> T T So (~p V q) --> ~q is true. I am not sure what the problem is, I'm guessing the first line. But your conclusion is not correct, because you did not test every case

51. Esmy

Oh lol thats just one of them lol. I did so many that I got lost ... but that I found out wasn't correct after all the thinking ;l

52. phi

good, sounds like you have a handle on it.