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Math Help - Formalizing a problem...

  1. #1
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    Formalizing a problem...

    Hello all, I have the following problem which I am attempting to formalize:

    Code:
    There are three containers, one gold, one silver, and one bronze. Inside one of these containers is an apple. The containers contain the following inscriptions:
    
    Gold Container: The apple is in this container
    Silver Container: The apple is not in this container
    Bronze Casket: At most one of the containers has a true inscription
    Ok, now I am trying to formalize...So far, I have:

    gc: Apple is in the gold container
    sc: Apple is in the silver container
    bc: Apple is in the bronze container

    g: Apple is in the gold container (inscription)
    s: Apple is not in the silver container (inscription)
    b: At most, one of g, s, or b is true (inscription)

    F0: g ~ gc
    F1: s ~ !sc
    F2: ?

    (where ~ means equivalence, and ! means NOT)

    What would F2 become (how would I translate b?) I dont know how I would write it, the "at most" is confusing me. I need F2, as this will most likely be the statement that I will simplify in order to derive the answer...

    Any ideas? Thanks!
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  2. #2
    ddt
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    Quote Originally Posted by inthedl View Post
    g: Apple is in the gold container (inscription)
    s: Apple is not in the silver container (inscription)
    b: At most, one of g, s, or b is true (inscription)

    F0: g ~ gc
    F1: s ~ !sc
    F2: ?

    (where ~ means equivalence, and ! means NOT)

    What would F2 become (how would I translate b?) I dont know how I would write it, the "at most" is confusing me. I need F2, as this will most likely be the statement that I will simplify in order to derive the answer...

    "At most one" means "zero or one", so you get

    b ~ (!g ^ !s ^ !b) V (g ^ !s ^ !b) V (!g ^ s ^ !b) V (!g ^ !s ^ b)

    and I guess you can take it from there. Of course that formula needs some simplifying.
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  3. #3
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    Thanks, I understand what "at most" is referring to now. Can I simplify that formula by using...

    (g XOR s XOR b) OR (!g ^ !s ^ !b) ?
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  4. #4
    ddt
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    Quote Originally Posted by inthedl View Post
    Thanks, I understand what "at most" is referring to now. Can I simplify that formula by using...

    (g XOR s XOR b) OR (!g ^ !s ^ !b) ?
    No, the part (g XOR s XOR b) is also true if all three g, s, and b are true.
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  5. #5
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    Ok, so how about this...

    b is the same as (!g ^ !s ^ !b) V (g ^ !s ^ !b) V (!g ^ s ^ !b) V (!g ^ !s ^ b)
    So if b is true then so is (!g ^ !s ^ !b) V (g ^ !s ^ !b) V (!g ^ s ^ !b) V (!g ^ !s ^ b)
    So I can use
    AND here and say b AND [(!g ^ !s ^ !b) V (g ^ !s ^ !b) V (!g ^ s ^ !b) V (!g ^ !s ^ b)]
    And I can throw out any !b, since I can't have b and !b at the same time.
    So it simplifies to (!g ^ !s ^ b)

    Which, using "Distributivity over ^", would give me (!g ^ s) ^ (!g ^ b), So, the apple is both in the silver container and the bronze container, that cant be right...


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  6. #6
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    Or, if I stick with (!g ^ !s ^ b), Does that mean that it has to be in b? Or just that the Inscription on b is true?
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  7. #7
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    I think it has to, because s ~ !sc, so its not in the silver, and its not in the gold (!g)...
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  8. #8
    ddt
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    Quote Originally Posted by inthedl View Post
    Ok, so how about this...

    b is the same as (!g ^ !s ^ !b) V (g ^ !s ^ !b) V (!g ^ s ^ !b) V (!g ^ !s ^ b)
    So if b is true then so is (!g ^ !s ^ !b) V (g ^ !s ^ !b) V (!g ^ s ^ !b) V (!g ^ !s ^ b)
    So I can use
    AND here and say b AND [(!g ^ !s ^ !b) V (g ^ !s ^ !b) V (!g ^ s ^ !b) V (!g ^ !s ^ b)]

    And I can throw out any !b, since I can't have b and !b at the same time.

    So it simplifies to (!g ^ !s ^ b)

    Which, using "Distributivity over ^", would give me (!g ^ s) ^ (!g ^ b), So, the apple is both in the silver container and the bronze container, that cant be right...
    No, you're forgetting that both also can be false.

    In fact, you have the formula

    b iff P (where P is the long formula), or in other words:
    b <=> P, or written out in only NOT, AND and OR:
    (b ^ P) V (!b ^ !P)

    The first part, (b ^ P), is indeed equivalent to (!g ^ !s ^ b), but the second part is a bit more complicated. Use the De Morgan formula to simplify it, or just write out a truth table...
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  9. #9
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    Ok, after doing the truth tables, I found a discrepency that I cannot sort through...

    Code:
    g s b            (b^P)  (!g ^ !s ^ b)
    T T T             F              F
    T T F             F              F
    T F T             F              F
    T F F             F              F
    F T T             F              F
    F T F             F              F
    F F T             F              T
    F F F             F               F

    These statements should be equivalent but, for some reason, they are not.


    Assuming the T (where g,s,r := F,F,T), then the bomb is in the silver casket. It is not in the gold (!g), and it is in the silver because if s !sc, then !s sc. So, b is true. At most, one of g, s, or b has a true inscription – b has the true inscription, and g and s do not. Portia’s suitor should not open the silver casket. How come there is the discrepency?


    Here is the 2nd case, where (!b ^ !P):

    Code:
     
    g s b           (!b^!P) 
    T T T             F 
    T T F             T 
    T F T             F 
    T F F             T 
    F T T             F 
    F T F             T 
    F F T             F 
    F F F             F 
    



    I am confused as to how I can use this to help me figure out what
    (!b ^ !P) is equivalent to...I know that DeMorgan's would give me
    !(b V P)...
    Last edited by inthedl; October 19th 2008 at 10:36 PM.
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  10. #10
    ddt
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    Quote Originally Posted by inthedl View Post
    Ok, after doing the truth tables, I found a discrepency that I cannot sort through...

    Code:
    g s b            (b^P)  (!g ^ !s ^ b)
    T T T             F              F
    T T F             F              F
    T F T             F              F
    T F F             F              F
    F T T             F              F
    F T F             F              F
    F F T             F              T
    F F F             F               F
    These statements should be equivalent but, for some reason, they are not.
    With P, I meant the long formula P := (!g^!s^!b) V (g^!s^!b) V (!g^s^!b) V (!g^!s^b. It seems you just made mistake. See my truth table below.

    Quote Originally Posted by inthedl View Post
    Here is the 2nd case, where (!b ^ !P):
    Code:
    g s b           (!b^!P) 
    T T T             F 
    T T F             T 
    T F T             F 
    T F F             T 
    F T T             F 
    F T F             T 
    F F T             F 
    F F F             F 
    

    Again, some mistakes. My truth table for the whole lot:

    Code:
     g s b     P        (b^P)       (!g ^ !s ^ b)   (!b ^ !P)   (b^P) V (!bV!P)
     T T T     F        F              F                   F               F
     T T F     F        F              F                   T               T
     T F T     F        F              F                   F               F
     T F F     T        F              F                   F               F
     F T T     F        F              F                   F               F
     F T F     T        F              F                   F               F
     F F T     T        T              T                   F               T
     F F F     T        F              F                   F               F
    So the formula (!b ^ !P) boils down to (g ^ s ^ !b), and the whole formula boils down to (g ^ s ^ !b) V (!g ^ !s ^ b).

    Quote Originally Posted by inthedl View Post
    I am confused as to how I can use this to help me figure out what
    (!b ^ !P) is equivalent to...I know that DeMorgan's would give me
    !(b V P)...
    I suggested DeMorgan as an alternative to using truth tables. From the truth tables, you can just pick out the T's and try to combine them cleverly, without using DeMorgen. A handy tool for that is Karnaugh diagrams.
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  11. #11
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    ddt,

    I can't thank you enough for your help! You have really taught me how to appropriately approach these types of problems. So, the solution would be:

    (!g ^ !s ^ b) V (g ^ s ^ !b) – meaning it is either CASE 1, where the bomb is in the silver casket, or it is CASE 2, where the bomb is in the gold container(because s ~ !sc).
    But for CASE 2, would !b imply that it is not in the bronze container? Or just that the inscription on the bronze container is false?
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  12. #12
    ddt
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    Quote Originally Posted by inthedl View Post
    ddt,

    I can't thank you enough for your help! You have really taught me how to appropriately approach these types of problems. So, the solution would be:

    (!g ^ !s ^ b) V (g ^ s ^ !b) – meaning it is either CASE 1, where the bomb is in the silver casket, or it is CASE 2, where the bomb is in the gold container(because s ~ !sc).

    You're welcome! Indeed, that's the right conclusion.

    Quote Originally Posted by inthedl View Post
    But for CASE 2, would !b imply that it is not in the bronze container? Or just that the inscription on the bronze container is false?
    The latter: the inscription is false. The statement 'b' (the inscription) relates in no way to whether the apple is in the container or not. Note that the proposition 'bc' occurs nowhere in the argument you've set up.
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