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liliy Group Title

discrete math:(attached)

  • one year ago
  • one year ago

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  1. liliy Group Title
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    • one year ago
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  2. liliy Group Title
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    am i doing it right?

    • one year ago
  3. liliy Group Title
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    @mathmate ????

    • one year ago
  4. satellite73 Group Title
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    first of all the answer is definitely T

    • one year ago
  5. liliy Group Title
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    right... how?:(

    • one year ago
  6. satellite73 Group Title
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    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 \(p\lor\lnot q\)

    • one year ago
  7. mathmate Group Title
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    Remember that de Morgan's law requires you to interchange the and / or, like: ~(~a or b) <=> a ^ ~b

    • one year ago
  8. liliy Group Title
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    right...

    • one year ago
  9. mathmate Group Title
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    So there is a problem in the first term of the third line (second written line).

    • one year ago
  10. liliy Group Title
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    oy.

    • one year ago
  11. mathmate Group Title
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    The same problem with the second term. Otherwise, it will be the same as what you did.

    • one year ago
  12. mathmate Group Title
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    You skip a lot of steps, as @satellite73 pointed out. You may lose points in an exam.

    • one year ago
  13. satellite73 Group Title
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    \(\lnot (p\to q)\) \(\lnot (p\lor \lnot q)\) \(\lnot p \land q\) i believe

    • one year ago
  14. satellite73 Group Title
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    @mathmate probably has a better explanation, but as he/she said take it slow

    • one year ago
  15. liliy Group Title
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    so can you write it out?

    • one year ago
  16. mathmate Group Title
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    I will follow your thoughts.

    • one year ago
  17. satellite73 Group Title
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    \[\lnot[(p\to q) \land (q\to r)]\lor (q\to r)\] may be better, yes?

    • one year ago
  18. liliy Group Title
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    hold up. im gonna write it all out and you guys tell me if its right

    • one year ago
  19. satellite73 Group Title
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    but i kind of suck at this, so i will let mathmate check steps

    • one year ago
  20. mathmate Group Title
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    First replace all p->q etc with the or equivalents, ~[(~p or q) ^ (~q or r) ] or (~p or r)

    • one year ago
  21. mathmate Group Title
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    Now apply de Morgan's ~(~p or q) or ~(~q or r) or (~p or r)

    • one year ago
  22. mathmate Group Title
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    Apply de Morgan again: (p ^ ~q) or (q ^ ~r) or ~p or r

    • one year ago
  23. mathmate Group Title
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    Rearrange (commutativity of "or") ~p or (p^~q) or r or (q^~r)

    • one year ago
  24. mathmate Group Title
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    Now use associativity: [~p or (p^~q)] or [r or (q^~r)]

    • one year ago
  25. mathmate Group Title
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    Would that be obvious now?

    • one year ago
  26. liliy Group Title
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    • one year ago
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  27. liliy Group Title
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    BAsically im going to assume things to be True or false and then go on from there...

    • one year ago
  28. mathmate Group Title
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    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.

    • one year ago
  29. satellite73 Group Title
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    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

    • one year ago
  30. satellite73 Group Title
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    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 \(T\)

    • one year ago
  31. satellite73 Group Title
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    it is hidden in the last line written by @mathmate

    • one year ago
  32. mathmate Group Title
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    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

    • one year ago
  33. satellite73 Group Title
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    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 \(\lnot p \lor q\)

    • one year ago
  34. satellite73 Group Title
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    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 \(p \lor \lnot p\)

    • one year ago
  35. liliy Group Title
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    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...

    • one year ago
  36. satellite73 Group Title
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    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

    • one year ago
  37. liliy Group Title
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    no, that's wat i did originally and then i realized that in the instructions my professor said specifically NOT to use tables :(

    • one year ago
  38. satellite73 Group Title
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    it would look like this \[\begin{array}{|c|c|c|c|c|c|c|c} P & Q & R & P\Rightarrow{}Q & Q\Rightarrow{}R & (P\Rightarrow{}Q)\land{}(Q\Rightarrow{}R) & P\Rightarrow{}R & ((P\Rightarrow{}Q)\land{}(Q\Rightarrow{}R))\Rightarrow{}(P\Rightarrow{}R) \\ \hline 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 1 & 0 & 0 & 1 & 1 \\ 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ \hline 1 & 0 & 0 & 0 & 1 & 0 & 0 & 1 \\ 1 & 0 & 1 & 0 & 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 & 0 & 0 & 0 & 1 \\ 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ \hline \end{array}\]

    • one year ago
  39. satellite73 Group Title
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    well not really, i used a machine to do this

    • one year ago
  40. satellite73 Group Title
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    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

    • one year ago
  41. satellite73 Group Title
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    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

    • one year ago
  42. mathmate Group Title
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    Right you are!

    • one year ago
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is replying to Can someone tell me what button the professor is hitting...

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