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anonymous

  • one year ago

Please help me to Validate the argument by rules of inference ~R Q => R P V Q Then P ^ ~Q

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  1. anonymous
    • one year ago
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    Rule of inference

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  2. anonymous
    • one year ago
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    dunno how to put this to words, but with ->, see: http://www.millersville.edu/~bikenaga/math-proof/truth-tables/truth-tables13.png T F -> F F F -> T we have ~R and Q=>R given which means Q is also false the only way we can have P V Q is if P is true I'm not quite sure what I'm doing is right

  3. anonymous
    • one year ago
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    It valid (if solve by truth table) but I try to use by rules of inference, I can't solve i wanna know about solution to solve it

  4. anonymous
    • one year ago
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    ~R Q => R is modus tollens

  5. anonymous
    • one year ago
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    so now we have modus tollens ~R Q => R .:.~Q ----------------- P V Q Then P ^ ~Q --------------------------- with elimination, we have ~Q P V Q .:. P

  6. anonymous
    • one year ago
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    sorry that's so disorganized

  7. anonymous
    • one year ago
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    and what we will do with P ^ ~Q, how I know it is true

  8. anonymous
    • one year ago
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    I don't think there's an argument for it, it's just what we found we got P from elimination and ~Q from modus tolens

  9. anonymous
    • one year ago
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    Then we can use P and ~Q with conjuction ?

  10. anonymous
    • one year ago
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    oh, wow I'm blind yeah

  11. anonymous
    • one year ago
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    I feel confuse about definition of Propositional Logic that can use ~Q again

  12. anonymous
    • one year ago
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    do you understand why modus tollens is true logically?

  13. anonymous
    • one year ago
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    I dunno I use follow the rule

  14. anonymous
    • one year ago
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    no point in following rules blindly. this is the truth table we're working with\( \begin{array}{l|c|r} \text{P} & \text{Q} & \text{P}\implies \text{Q}\\ \hline 0 & 0 & 1 \\ 0 & 1 & 1\\ 1 & 0 & 0 \\ 1 & 1 & 1 \end{array} \)

  15. anonymous
    • one year ago
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    modus tollens states that \( p\implies q\\ \text{~}q\\\text{.^.~p} \)

  16. anonymous
    • one year ago
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    I think I have more understand it same modus tollens that you said from p => q is T and ~q is T then q = F then if p => q will T and p = F then ~p = T

  17. anonymous
    • one year ago
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    when we're given a statement, we assume it's true the ~ is a negation so p=>q means we only look at the cases where p=>q evaluates to true

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  18. anonymous
    • one year ago
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    then we look at the next statement ~q we need to find a place where q is 0

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  19. anonymous
    • one year ago
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    the conclusion from modus tollens says we have ~p or p=false

  20. anonymous
    • one year ago
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    and we can see it's true in the truth table

  21. anonymous
    • one year ago
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    Thanks a lot! I quite understand it

  22. anonymous
    • one year ago
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    no problem, what class is this for btw

  23. anonymous
    • one year ago
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    Discrete Math

  24. anonymous
    • one year ago
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    ah, same. I assume you're a fellow CS major

  25. anonymous
    • one year ago
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    out of curiosity, how much calc did you have to take

  26. anonymous
    • one year ago
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    I have grade B calc since freshman until second year

  27. anonymous
    • one year ago
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    I'm sorry if i slow to reply I'm not strong in English language, I try to use it more

  28. anonymous
    • one year ago
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    that's the best way to learn a language :) anyhow good luck with your class

  29. anonymous
    • one year ago
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    You too, Good luck and Thanks for everything that you gave me today.

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