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Math Help - Predicate logic - Translate?

  1. #1
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    Predicate logic - Translate?

    Where:

    Ax = x is diamond
    Gx = x shines
    Lxy = x likes y

    1. Not everything that shines is diamonds
    2. Someone likes diamonds
    3. Everyone who likes diamonds likes everything that shines
    4. Someone likes only thing that shines
    5. Everything that shines are liked by someone
    6. There is something that is neither diamond or shines, but are liked by everyone

    How would this logic expression look like....

    Appreciate any help I can get!
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  2. #2
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    Predicate Logic

    Hello jokke22
    Quote Originally Posted by jokke22 View Post
    Where:

    Ax = x is diamond
    Gx = x shines
    Lxy = x likes y

    1. Not everything that shines is diamonds
    You can do this in two ways: Either re-write it as 'It is not true that if x shines, then x is a diamond'; or as 'There exists an x, x shines and x is not a diamond'

    The first gives you: \neg (\forall x, Gx \Rightarrow Ax)

    and the second: \exists x, (Gx \wedge \neg Ax)

    Either of these is equally correct.

    2. Someone likes diamonds
    Re-write this as 'There exists an x, such that for all y, if y is a diamond then x likes y'

    So this symbolises as: \exists x, \forall y, Ay \Rightarrow Lxy

    3. Everyone who likes diamonds likes everything that shines
    This can be re-written 'If x is a diamond, and y likes x then if z shines, then y likes z'. I'll start you off:

    \forall x, y, (Ax \wedge Lyx) \Rightarrow \dots

    Can you finish it?

    4. Someone likes only thing that shines
    Re-write this as 'There exists an x, such that if x likes y, then y shines'

    5. Everything that shines are liked by someone
    This is 'There exists an x, such that if y shines, then x likes y.'

    6. There is something that is neither diamond or shines, but are liked by everyone
    'There exists an x, such that x is not a diamond and x does not shine and for all y, y likes x.'

    Can you complete these now?

    Grandad
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  3. #3
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    Thank you for your answer/enlightenment!

    Though, I have given it some tries now but im pretty green at the area here is what I've come up with...

    4. Someone likes only thing that shines

    \exists x, (Ax \wedge Gx) \Rightarrow Lxy

    5. Everything that shines is are liked by someone

    \exists x, (\forall x, Gx) \Rightarrow Lxy

    6. There is something that is neither diamond or shines, but are liked by everyone.

    \exists x, \neg (\forall x \wedge Gx) \Rightarrow Lxy


    I think i'm way off but... If you/anyone have the time please elaborate!
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  4. #4
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    Predicate Logic

    Hello jokke22
    Quote Originally Posted by jokke22 View Post
    Thank you for your answer/enlightenment!

    Though, I have given it some tries now but im pretty green at the area here is what I've come up with...

    4. Someone likes only thing that shines

    \exists x, (Ax \wedge Gx) \Rightarrow Lxy
    You have this the wrong way round. y is the thing that shines; and this is if x likes y. So it is:

    \exists x, \forall y, Lxy \Rightarrow Gy

    5. Everything that shines is are liked by someone
    \exists x, (\forall x, Gx) \Rightarrow Lxy
    Again, y is the thing that shines, so it is:

    \exists x, \forall y, Gy \Rightarrow Lxy

    6.
    There is something that is neither diamond or shines, but are liked by everyone.

    \exists x, \neg (\forall x \wedge Gx) \Rightarrow Lxy
    Note there are three propositions connected by 'and'. They are:

    • x is not a diamond
    • x does not shine
    • y likes x

    So it is:

    \exists x, \forall y, (\neg Ax) \wedge (\neg Gx) \wedge (Lyx)

    Grandad
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  5. #5
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    Quote Originally Posted by Grandad View Post
    Hello jokke22You have this the wrong way round. y is the thing that shines; and this is if x likes y. So it is:

    \exists x, \forall y, Lxy \Rightarrow Gy

    Again, y is the thing that shines, so it is:

    \exists x, \forall y, Gy \Rightarrow Lxy

    Note there are three propositions connected by 'and'. They are:

    • x is not a diamond
    • x does not shine
    • y likes x

    So it is:

    \exists x, \forall y, (\neg Ax) \wedge (\neg Gx) \wedge (Lyx)

    Grandad
    Thanks a million Grandad! Appriciate the help!

    I have to look it over once more to fully understand how it all hangs together but this helps a great deal!
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