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Convert statements into first order logic

  1. Jan 6, 2017 #1
    Given are the following predicate symbols:
    Member(x) : x is a member of the bicycle club
    Chairman(x) : x is the chairman of the bicycle club
    Bicycle(x) : x is a bicycle
    Brand(x, y) : the brand of x is y
    Owns(x, y) : x is the owner of y

    a. Statement: every member of the bicycle club has the same brand of bicycle.

    Note that this statement is ambiguous, the meaning could be:
    i: each member of the Club has bycicles of the same brand (monobrand member),
    ii: all the members of the Club have bycicles of the same brand (monobrand Club),
    iii: all the members of the Club have at least one bicycle of the same brand, but (s)he may
    have other bicycles of other brands.

    Express these interpretations of the given statement in first order logic, using the predicate
    symbols above.

    I'm trying for example the first one but i'm having some problems:
    i. ∀x ∀y (member(x) ∧ bicycle(y) ∧ owns(x,y) ∃ z brand(y,z)¬∃ z2 brand(y,z2))
    Or maybe it's better this:
    i. ∀x ∀y (member(x) ∧ bicycle(y) ∧ owns(x,y) → ¬∃ z,z2 (brand(y,z2) ∧ brand(y,z)))

    There is a correct answer or there could be an easier way to do this? I'll keep update this post if i will solve it. Thanks :wink:
     
    Last edited: Jan 6, 2017
  2. jcsd
  3. Jan 6, 2017 #2

    stevendaryl

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    Both of those are sort of close, but not exactly. Look at the first one: (I've rewritten it in LaTex):
    [itex]\forall x \forall y (member(x) \wedge bicycle(y) \wedge owns(x,y) \wedge \exists z\ brand(y,z) \rightarrow \neg \exists z_2\ brand(y,z_2))[/itex]

    The clause--[itex]\neg \exists z_2\ brand(y,z_2)[/itex]--literally says that "y does not have a brand". What you need to say, instead is that "y does not have a brand that is different from z".
     
  4. Jan 6, 2017 #3
    Thank you for the answer, i'm trying to translate that "different from z" but i'm not finding the right way to do that in FOL, the following statement is wrong too?
    ∀x ∀y (member(x) ∧ bicycle(y) ∧ owns(x,y) ∃ z brand(y,z)¬∃ z2 (brand(y,z2) ∧ brand(y,z)))

    It's probably wrong because it's saying that for each bicycle not exist more than one brand associated with it...
     
    Last edited: Jan 6, 2017
  5. Jan 6, 2017 #4

    Mark44

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    @Marclan, in future posts, please don't delete the three parts of the homework template -- they are required here.
     
  6. Jan 7, 2017 #5
    Still trying to understand how to convert the first sentence, can anyone give an advice to help me?
    The problem is translating this correct clue "y does not have a brand that is different from z".
     
  7. Jan 8, 2017 #6
    In this moment i'm thinking that it's not possible to say that in the first order logic. I'm still on my solution of the third post.
     
  8. Jan 9, 2017 #7
    Can anyone check if this can be a solution for the second statement? Thanks ;)
    ∀x ∀y ∀z ∀x2 ∀y2 (member(x) ∧ bicycle(y) ∧ owns(x,y) ∧ brand(y,z)member(x2) ∧ bicycle(y2) ∧ owns(x2,y2) ∧ brand(y2,z))

    I don't know if it's more correct use ∃z or ∀z.
     
  9. Jan 9, 2017 #8

    PeroK

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    This isn't my area, but it terms of normal mathematical logic, for (i) I would start:

    For all ##x## there exists ##z## ##\dots##

    where ##x## is a member and ##z## (depending on ##x##) is the brand of bicycle that ##x## has exclusively.
     
  10. Jan 9, 2017 #9
    I wrote the iii) and i think it's correct:
    ∀x ∀y ∀z (member(x) ∧ bicycle(y) ∧ owns(x,y) ∧ brand(y,z)∀x2 ∃y2 (member(x2) ∧ bicycle(y2) ∧ owns(x2,y2) ∧ brand(y2,z)))

    Always here trying to resolve i) and ii) and learn something new. Thanks!
     
  11. Jan 9, 2017 #10

    PeroK

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    That doesn't look right at all. (iii) ought to start with: there exists a brand ##z## ...
     
  12. Jan 9, 2017 #11

    PeroK

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    @Marclan have you done any (mathematical) logic before? My approach would be to translate the English into logic, then (what I can't do) translate the logic into the first-order mumbo-jumbo. But, if you can't do "normal" logic, it's going to be hard to go straight to the first-order expressions.

    For example the phrase I've underlined here is logically superfluous:

    iii: all the members of the Club have at least one bicycle of the same brand, but (s)he may
    have other bicycles of other brands.
     
  13. Jan 9, 2017 #12
    It's not so simple in the first order logic. Can you explain me why my last statement is wrong? Translating that in words sounds like: "For each member who has a bicycle of a brand z, there is for each other member (x2) at least one bicycle Y2 of the same brand z. It's not what iii) means?
     
  14. Jan 9, 2017 #13
    Yes i saw that too, i'm just ignoring it.
     
  15. Jan 9, 2017 #14

    PeroK

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    That's not what (iii) means. (iii) means there is some brand (call it the club's default brand) and all members have at least one bicycle of that brand.
     
  16. Jan 9, 2017 #15

    PeroK

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    It's clearly wrong because (iii) says nothing about "all brands" that are owned by at least one club member, which is the gist of what you have.

    I'd start again and try at least to get a clear logical statement of what (i)-(iii) mean. Without that you will just go round in circles.
     
  17. Jan 9, 2017 #16
    I'm sure about a thing, in the text there isn't a specific brand for the Club (instead of how it could be normal in the reality) so i'm thinking that a right way to solve it is using a general brand. And i don't see the problem "all the brands" because in a normal instantiation of the bicycles, each one will have only one brand so i'm just only saying for each member, each bicycle of him/her and for each brand associated (one) etc...
    For example, i can change that in the following statement (exist instead of each for z) but i don't see the advantage:

    ∀x ∀y z (member(x) ∧ bicycle(y) ∧ owns(x,y) ∧ brand(y,z)∀x2 ∃y2 (member(x2) ∧ bicycle(y2) ∧ owns(x2,y2) ∧ brand(y2,z)))

    Thank you for the support!
     
  18. Jan 9, 2017 #17

    PeroK

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    I still think you're going wrong on both the logical interpretation of the statement and then the rendering of that statement into first-order logic.

    For example, the ##\exists \ z ## is superfluous and what you have means something like:

    If a club member owns a bicycle of brand z, then all members of the club own a bicycle of brand z.
     
  19. Jan 9, 2017 #18
    You are right, it's the correct meaning. My statement doesn't say that "at least one". In this moment i don't really know how to solve all of these. Thanks anyway.
    Always opened for new ideas ;)
     
  20. Jan 9, 2017 #19

    PeroK

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    I'm going offline now anyway. But, my advice is you have to develop your logical thinking. Here's how I look at statements (i)-(iii). I do three things: try to figure out what it means (obviously!); think of an example of when it is true; figure out what it doesn't mean; think of an example of when it is false.

    (i) each member of the Club has bicycles of the same brand (monobrand member)

    What does it mean? It means every club member has his/her own preferred brand and all his/her bikes are of that brand.

    Example: Three members: A has one bike (brand X); B has 6 bikes (all brand Y); C has two bikes (all brand X)

    What does it not mean? It doesn't mean there is any relationship between the brands of bicycles preferred by different members.

    What else does it mean? It means no one can have two bikes of different brands.

    Example: A has two bikes, one brand X, one brand Y. This is not allowed by statement (i).

    That's the way I would analyse this. And, at this point I believe I truly understand statement (i).
     
  21. Jan 9, 2017 #20
    A good help but i'm still on my past solution:

    ∀x ∀y (member(x) ∧ bicycle(y) ∧ owns(x,y) ∃ z brand(y,z)¬∃ z2 (brand(y,z2) ∧ brand(y,z)))

    And i have just wrote that it's probably wrong because it's only saying that for each bicycle not exist more than one brand associated with it. If only i could know how to write this, i will solve all the exercize.
     
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