Help with some formal logic questions (Quantifiers and Predicates)

Discussion in 'Mac Programming' started by 103734, Apr 11, 2010.

  1. 103734 Guest

    Joined:
    Apr 10, 2007
    #1
    Ok I am having a hard time understanding this.

    If P(x) is “x is a person” and Q(x) be “x is nice”, than
    (∀x)[P(x) → Q(x)]
    Could be translated as "For all things, if it is a person than it is nice" or "Each person is nice".

    I get that part, but I don't get why
    (∃x)[P(x) → Q(x)]
    Doesn't translate as "For some things, if it is a person than it is nice", and why the statement is false.
     
  2. telecomm macrumors 65816

    telecomm

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    #2
    Well, the second translation would be grammatically incorrect ("some things" vs "it" - either both should be plural or both singular). I'm guessing that's not what you mean, though. ;)

    Existentially quantified conditional expressions can be a bit tricky. The truth of (∃x)[P(x) → Q(x)] requires the existence of an object, call it a, such that if a is a person, then a is nice. To see why this sentence is not equivalent to "Some people are nice" (which, I think, is what you were getting at in your question), think about each sentence's commitment to the existence of people.
     
  3. 103734 thread starter Guest

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    Apr 10, 2007
    #3
    Yea thats pretty much what I am trying to figure out.

    So it translates to "If a is a person then a is nice."? I don't get why this isn't a correct translation for "Some people are nice".

    Now that I think of it is it because the truth table for A→B? Because if A is false than it automatically makes the statement true?
     
  4. telecomm macrumors 65816

    telecomm

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    #4
    It's really tough to try to help with this without just giving away the answers. :eek: (I'm guessing this is homework, right?)

    That isn't quite the right translation—you can generally translate "∃x" directly as "There is a thing x such that..."

    That's the key. If you claim "Some people are nice" you're committing yourself to the existence of a certain sort of object, namely a nice person (and so the connective would be... something other than a conditional ;)). This isn't quite what's required for the truth of (∃x)[P(x) → Q(x)]. Again, I hesitate to say too much more, but you're on the right track. :)
     
  5. Cromulent macrumors 603

    Cromulent

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    #5
    This is a guess, I'm still learning this myself but it would be appear that "if a is a person, then a is nice" could be rewritten as "any person is nice" or "every person is nice". Therefore some people are nice is wrong because if a is not a person then they are not nice, but if they are a person then they are nice.

    "Some people are nice" suggests there are people who are not nice, which the first statement does not allow.

    Edit: Hah, I guess I don't know what I am talking about. Feel free to ignore me :).
     
  6. telecomm macrumors 65816

    telecomm

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    #6
    This is right (but we were translating a different sentence, and translating in the other direction, as you noticed). ;)

    You want to be careful here—it's better to ignore "what people usually mean to suggest when they say", etc., and consider only what is "strictly speaking" required for their claims to be true. So, while people usually aim to be "maximally informative" (and so might avoid saying things like "Some people are nice" unless there are people who are not nice), in logic you're only going to focus on what's relevant to the truth/falsity of a statement.
     
  7. naples98 macrumors member

    naples98

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    #7
    ∀x is the universal quantifier and can be translated as "For all instances of x" some property holds true. For example, all instances of cats are animals.


    ∃x is the existential quantifier and can be translated as "There exists at least one instance of x, such that" some property holds true. For example, there exists at least one animal in the ocean that is a mammal (e.g. dolphin).

    I'm not sure if this helps with your problem but I wanted to make sure you understood the difference between the two quantifiers.
     
  8. 103734 thread starter Guest

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    Apr 10, 2007
    #8
    lol, this isn't for homework but it is for school. The homework is pretty straight forward but I seen this in the notes and wanted to make sure I understand it before the tests.

    But just to make sure I got this, some people are nice doesn't translate to (∃x)[P(x) → Q(x)] because with the same inputs they return different results, like if a person doesn't exist than "some people are nice" it would return false, while (∃x)[P(x) → Q(x)] returns true, ect.

    Thanks.
     
  9. telecomm macrumors 65816

    telecomm

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    #9
    Yeah, that's about right. Instead of phrasing things in terms of input/output, it might be easier to think about the "way the world would have to be" in order for each statement to come out true. So, because the conditional comes out true with a false antecedent, (∃x)[P(x) → Q(x)] could be vacuously true in a world, i.e., true simply because there aren't any people. The same situation would be one in which "some people are nice" comes out false, since it would be represented using the given predicates as (∃x)[P(x) ∧ Q(x)], and so is true iff there is (at least one) nice person. In a world without people it would have to be false.

    Make sense?
     
  10. chown33 macrumors 604

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    Aug 9, 2009
    #10
    Impossible to read that without hearing it in his voice:
    http://en.wikipedia.org/wiki/Don_LaFontaine
     
  11. 103734 thread starter Guest

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    Apr 10, 2007
    #11
    Yea I am pretty sure I get it now, thanks a lot.
     

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