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Math Help - propositional logic

  1. #1
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    propositional logic

    Hi, Can someone take a look if I have worked through this correctly? thanks

    Question 4.
    ~r -> p, (~p) v q, ~s -> (~p)^(~r), (~p) ^ r -> (~s) v t, ~q, therefore t.

    1.~q premise
    ~q vp premise
    therefore ~p disjunctive syllogism

    2.~p from 1
    ~r -> p premise
    therefore r modus tollens

    3.~p from 1
    r from 2
    therefore ~p^r conjunctive addition

    4.~p^r from 3
    ~s -> (~p)^(~r) premise
    therefore s modus tollens

    5.~p^r from 3
    s from 4
    (~p) ^ r -> (~s) v t therefore t disjunctive syllogism

    thanks
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  2. #2
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    (4) is an incorrect application of modus tollens.

    From ~s->(~p^~r), to get s by modus tollens, you need to have ~(~p^~r). Which rules of inference are you using? Can you use De Morgan's Law? Also, in (1), I think you mean, ~p v q, instead of ~q v p. And finally, in step 5, you seem to be compressing to steps into one.
    Last edited by quiney; August 3rd 2010 at 12:42 AM.
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  3. #3
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    I don't know about others, but a lot of your OP is gibberish in my browser. Any chance you could re-type it using LaTeX? You can use \neg for negation, \vee for or, \land for and, \to for implication, etc.
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  4. #4
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    Thanks quiney

    So in 4
    can I use -p^r as a premise?
    -p^r is same as -(-rvp) DeMorgans


    5. -rvp, -s --> -(-rvp) therefore s Modus tollens

    ???
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  5. #5
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    Since you already have r, try using disjunctive addition to get (pvr). Then use De Morgan's and see what you can get from there.
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  6. #6
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    \neg r \to p, (\neg p) \vee q, \neg s \to (\neg p) \land (\neg r), (\neg p) \land r \to (\neg s) \vee t, \neg q

    (  \neg  \vee  \land  \to) <- my reference**

    1. \neg q premise
    \neg p \vee q premise
    \therefore \neg p disjunctive syllogism

    2. \neg p from 1
    \neg r \to p premise
    \therefore r modus tollens

    3. \neg p from 1
    r from 2
    \therefore \neg p \land r conjunctive addition

    4.r from 2
    \therefore p \vee r

    5. p \vee r from 3
    I don't know how to use DeMorgans here to get (\neg p) \land (\neg r)
    I understand DeMorgans to be (\neg p) \land (\neg r) \equiv \neg (p \vee q)

    ??
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  7. #7
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    Thanks much for the clarification. It reads much better now.

    A couple of comments.

    1. Your step 5 should reference step 4, not step 3.
    2. I'm confused as to what exactly your premisses are, and what your desired conclusion is. If you're trying to show \neg p\land\neg r from the following premisses:

    \neg r\to p
    \neg p\vee q
    \neg q,

    then I think your proof has just managed to disprove it. Why? Because you've proven \neg p, as well as r. Therefore, \neg p\land \neg r is false.
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  8. #8
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    Quote Originally Posted by dunsta View Post

    5. p \vee r from 3
    I don't know how to use DeMorgans here to get (\neg p) \land (\neg r)
    I understand DeMorgans to be (\neg p) \land (\neg r) \equiv \neg (p \vee q)
    You're not trying to get ~p & ~r, you're trying to get ~(~p & ~r). (p v r) is equivalent to ~(~p & ~r). Remember, De Morgan's works both ways.

    Then, use modus tollens and your third premise and you're almost done.
    Last edited by quiney; August 3rd 2010 at 05:05 AM.
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  9. #9
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    \neg r \to p, (\neg p) \vee q, \neg s \to (\neg p) \land (\neg r), (\neg p) \land r \to (\neg s) \vee t, \neg q, \therefore t

    I'm trying to prove t

    I can see from here (\neg p) \land r \to (\neg s) \vee t, \neg q, that (\neg s) \vee t means s is true, then -s or t, results in t is true i.e. proving the conclusion.

    I too am confused as to how to get there using the basic inference rules, Ackbeet.

    I think step 5 needs to show that s is true using \neg s \to (\neg p) \land (\neg r) then I can go on to use the premise from 3 with (\neg p) \land r \to (\neg s) \vee t to show t,

    which is what Im working on now with

    5. p \vee r from 4
    p \vee r \equiv \neg(\neg p \land \neg r) DeMorgans

    6. \neg(\neg p \land \neg r) from 5
    \neg s \to (\neg p) \land (\neg r)
    \therefore s modus tollens

    7. \neg(p) \land r from 3
    s from 6
    \neg s \vee t premise
    therefore t disjunctive syllogism

    ??
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  10. #10
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    \neg r \to p, (\neg p) \vee q, \neg s \to (\neg p) \land (\neg r), (\neg p) \land r \to (\neg s) \vee t, \neg q, \therefore t

    I'm trying to prove t

    I can see from here (\neg p) \land r \to (\neg s) \vee t that (\neg s) \vee t means s is true, then -s or t, results in t is true i.e. proving the conclusion.

    I too am confused as to how to get there using the basic inference rules, Ackbeet.

    I think step 5 needs to show that s is true using \neg s \to (\neg p) \land (\neg r) then I can go on to use the premise from 3 with (\neg p) \land r \to (\neg s) \vee t to show t,

    which is what Im working on now with

    5. p \vee r from 4
    p \vee r \equiv \neg(\neg p \land \neg r) DeMorgans

    6. \neg(\neg p \land \neg r) from 5
    \neg s \to (\neg p) \land (\neg r)
    \therefore s modus tollens

    7. \neg(p) \land r from 3
    s from 6
    \neg s \vee t premise
    \therefore t disjunctive syllogism

    ??
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  11. #11
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    You've got it. Just clean it up a bit, and, if you wouldn't mind, post the entire proof like this:

    Premiss 1
    Premiss 2
    ...
    -------------
    1. \neg q
    etc.

    n. t, from ____.

    Make sense?
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  12. #12
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    Thanks Ackbeet and quiney, I really mean that.
    As it's late here, and I can sleep well now this is solved, I will post the full solution 2mw if you like Ackbeet.

    Thanks for taking the time to help.
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  13. #13
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    You're welcome for whatever help I was. I do think posting the entire solution will do you good. You'll be able to see better the flow of the whole solution.
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