Page 1 of 2 12 LastLast
Results 1 to 15 of 21

Math Help - Logical statements

  1. #1
    Senior Member
    Joined
    Apr 2009
    Posts
    306

    Logical statements

    Peter states the following facts about the football premiership:
    If St Kilda beat Hawthorn, Franklin is included next week
    Hawthorn beating St Kilda means that Geelong will top the ladder
    Hawthorn will make the top eight only if Carlton miss out on the finals
    If Geelong lose to Essendon then Geelong can't top the ladder
    If Hawthorn fail to make the top eight, Franklin will be dropped next week

    a) Formulate these five facts as statements in propositional logic.

    I did this:

    p: St Kilda beat Hawthorn
    q: Franklin included next week
    x: Geelong top the ladder
    y: Hawthorn will make the top 8
    z: Carlton miss out on finals
    a: Geelong lose to Essendon

    1. p \to q
    2. \neg p \to x
    3. y \leftrightarrow z
    4. a \to \neg x
    5. \neg y \to \neg q

    b) Is it the case that if Geelong don't manage to top the ladder, Carlton miss out
    on the finals? Give your reasoning.

    So is this part asking whether \neg x \to z is true or false?

    Do I use rules of inference from my 5 propositions from part a) to figure out b)? I can't seem to link them together...

    Any help would be much appreciated!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Mar 2009
    Posts
    378
    Deleted
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Apr 2009
    Posts
    306
    ?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Mar 2009
    Posts
    378
    I posted something, but I decided it had been too long since I have done this and wasn't confident in the solution so I deleted it.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Apr 2009
    Posts
    306
    Ohhh right, haha, sorry, I thought you meant I had to delete my post lolz
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Apr 2009
    Posts
    306
    Ok I 'think' I got it after many trials, can anyone confirm if my working is correct/wrong. Thanks very much!

    From 1. p \to q \equiv \neg q \to \neg p name this 1'.
    Assuming 5 and 1' are true and then using Hypothetical Syllogism we yield \neg y \to \neg p \equiv p \to y to be true.

    From 3. y \leftrightarrow z \equiv (y \to z) \land (z \to y) name this 3'. Assuming 3' is true then using Simplification yields y \to z to be true.

    Looking at p \to y and y \to z using Hypothetical Syllogism we yield p \to z to be true.

    From 2. we have \neg p \to x \equiv \neg x \to p name this 2'.

    Using Hypothetical Syllogism with 2' and p \to z yields \neg x \to z.

    Thus \neg x \to z is true!
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member
    Joined
    Mar 2009
    Posts
    378
    I guess I could have left my answer, but yours is prettier.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Referring to the OP, your statement 3 is incorrect. It should be

    z\to y.

    The "only if" keyword works like this: A only if B gets translated B\to A. The bidirectional arrow is the translation of "if and only if".
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1
    Quote Originally Posted by Ackbeet View Post
    The "only if" keyword works like this: A only if B gets translated B\to A.

    The bit above in red is incorrect.
    It should be: \text{A only if B} \equiv \left( {A \to B} \right).

    Reason: It says that A is true only if B is also true.
    The only condition we cannot allow is true implies false.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Senior Member
    Joined
    Apr 2009
    Posts
    306
    Thanks for the help guys!

    Ackbeet, thanks for that pickup, that was a stupid mistake by me - however I was under the impression that "A if B" means B implies A, however "A only if B" means A implies B?

    edit - ah just as I posted this, plato confirmed it haha
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Senior Member
    Joined
    Mar 2009
    Posts
    378
    Quote Originally Posted by usagi_killer View Post
    Thanks for the help guys!

    Ackbeet, thanks for that pickup, that was a stupid mistake by me - however I was under the impression that "A if B" means B implies A, however "A only if B" means A implies B?

    edit - ah just as I posted this, plato confirmed it haha
    Yes, because that is where if and only if comes from, i.e.,  A\to B \wedge B \to A \equiv A \leftrightarrow B
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Senior Member
    Joined
    Apr 2009
    Posts
    306
    Hmmm, it seems I am stuck with a further extension of this problem.

    The extension is that "can one deduce that either Geelong lose to Essendon or Carlton make the finals?"

    So my interpretation of this question is that is a \lor \neg z true or false?

    Now to prove this, if a is true or \neg z is true then a \lor \neg z is true, however if a and \neg z are both false, then a \lor \neg z is false.

    But no matter how hard I try, I can not seem to find any way of manipulating the information we have to prove what I described above
    Follow Math Help Forum on Facebook and Google+

  13. #13
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Reply to Plato: Yeah, I was thinking of "A if B". Sorry about that.
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Junior Member
    Joined
    Aug 2010
    Posts
    32
    Quote Originally Posted by usagi_killer View Post
    Hmmm, it seems I am stuck with a further extension of this problem.

    The extension is that "can one deduce that either Geelong lose to Essendon or Carlton make the finals?"

    So my interpretation of this question is that is a \lor \neg z true or false?

    Now to prove this, if a is true or \neg z is true then a \lor \neg z is true, however if a and \neg z are both false, then a \lor \neg z is false.

    But no matter how hard I try, I can not seem to find any way of manipulating the information we have to prove what I described above
    I think that inference is invalid.

    When P: T, Q: T, X: T, Y: T, Z: T, A: F, then the argument is such that all the premises are true, and the conclusion is false. Therefore, the argument is invalid. Use a truth table to check.
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Senior Member
    Joined
    Apr 2009
    Posts
    306
    Thanks for your help but I don't quite understand how to solve it yet.

    I did this:

    a \lor \neg z \equiv z \to a

    So now I have changed it to z \to a I don't know how to show whether that is true or false.
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Replies: 7
    Last Post: November 27th 2011, 11:39 AM
  2. [SOLVED] Confused about logical equivalence of some statements
    Posted in the Discrete Math Forum
    Replies: 8
    Last Post: July 13th 2011, 04:44 AM
  3. Replies: 1
    Last Post: July 8th 2011, 05:21 AM
  4. Prove/disprove using logical using logical arguments
    Posted in the Discrete Math Forum
    Replies: 5
    Last Post: February 24th 2010, 06:29 AM
  5. Replies: 3
    Last Post: January 21st 2010, 07:45 AM

Search Tags


/mathhelpforum @mathhelpforum