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Esmy

  • 2 years ago

Use truth tables to test the validity of the argument. p → ~q q → ~p ∴ p ∨ q

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  1. Esmy
    • 2 years ago
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    this is valid

  2. Esmy
    • 2 years ago
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    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
    • 2 years ago
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    All four cases come out T, so the argument is valid. I think lol.

  4. A.Avinash_Goutham
    • 2 years ago
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    u need to verify that using a truth table?

  5. Esmy
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    "p -> ~q" and "q -> ~p" the last row has T and T so the conclusion is false?

  8. A.Avinash_Goutham
    • 2 years ago
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    actuall it mean that u shud conside the cases on when both of them are true

  9. Esmy
    • 2 years ago
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    Oh I totally miss understood the question

  10. Esmy
    • 2 years ago
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    So how would you do that? :o

  11. Esmy
    • 2 years ago
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    help can you explain to how you do it :/

  12. A.Avinash_Goutham
    • 2 years ago
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    consider those case when p → ~q q → ~p are true find them

  13. Esmy
    • 2 years ago
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    can you explain like in steps :/ I understand in steps.

  14. A.Avinash_Goutham
    • 2 years ago
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    hmm consider those inputs for which p → ~q is true q → ~p is true

  15. Esmy
    • 2 years ago
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    umhmm :o

  16. A.Avinash_Goutham
    • 2 years ago
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    lol sorry....... u shud try and understand

  17. A.Avinash_Goutham
    • 2 years ago
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    ok fine q → ~p is true for first three columns in the table

  18. Esmy
    • 2 years ago
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    uhmm :o

  19. Esmy
    • 2 years ago
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    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
    • 2 years ago
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    Help D:

  21. A.Avinash_Goutham
    • 2 years ago
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    u dont need help gal jus think

  22. Esmy
    • 2 years ago
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    Blagh I am thats why I'm asking if I did right or not

  23. Esmy
    • 2 years ago
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    Okay I'm guessing i didn't do that one right so ill work on another one.

  24. A.Avinash_Goutham
    • 2 years ago
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    lol draw truth tables for p → ~q q → ~p and pvq see when p → ~q q → ~p are true pvq is true

  25. Esmy
    • 2 years ago
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    (~p V q) --> ~q (~T V F) --> ~F (F V F) --> T F --> T T So (~p V q) --> ~q is true.

  26. Esmy
    • 2 years ago
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    there

  27. Esmy
    • 2 years ago
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    Blagh W/e I'm right I've done everything... :l

  28. Esmy
    • 2 years ago
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    q p (q → ~p) → (q ∧ ~p) T T T T F T F T F F F F

  29. satellite73
    • 2 years ago
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    the first two statements are identical \[p\to \lnot q\iff q\to \lnot p\] one is the contrapositive of the other

  30. Esmy
    • 2 years ago
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    hmmm is my table okay :/ ?

  31. Esmy
    • 2 years ago
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    I've done like three so far -_-

  32. satellite73
    • 2 years ago
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    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
    • 2 years ago
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    write

  34. phi
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    so none of my tables where right :/ ...

  38. phi
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    of course not! maybe it is sunny and i stay home anyway!

  41. Esmy
    • 2 years ago
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    Oh okay :o... I c I c

  42. phi
    • 2 years ago
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    I don't see the other problems..... with premises and conclusion

  43. Esmy
    • 2 years ago
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    neither do I

  44. phi
    • 2 years ago
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    I mean, did you post them?

  45. Esmy
    • 2 years ago
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    yeah I did ?

  46. phi
    • 2 years ago
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    Is this one of the problems (~p V q) --> ~q ? with premise ~p v q and conclusion ~q

  47. Esmy
    • 2 years ago
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    I posted my answer and everything D:?

  48. phi
    • 2 years ago
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    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
    • 2 years ago
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    did you do this one or I did I forgot...

  50. phi
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    good, sounds like you have a handle on it.

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