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  1. #1
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    Modal logic / Possible world hijinx?

    So here goes.

    A little primer on modal logic.

    A quick-and-dirty explanation of logical operators:

    (p AND q) means p is true and q is true
    (p OR q) means p is true, q is true, or both p and q are true.
    ~p is a proposition that has the opposite truth value of p.

    USEFUL PROPERTIES

    ~~p = p

    DeMorgan's laws:
    ~(p AND q) = (~p OR ~q)

    I.e., if it's false that "John is tall AND Susie is tall", either John or Susie or both must not be tall.

    ~(p OR q) = (~p AND ~q)
    I.e., if it's false that "He ate cake OR he ate ice cream" then he couldn't have eaten cake AND he couldn't have eaten brownies.

    So you kind of "distribute" the ~ through by negating each proposition and flipping OR with AND (and vice versa). It's kind of ugly with nested operators (i.e., A AND (B OR (C AND D))), but it can be done by continuing to apply the rule.

    You can also do it in reverse:

    (p AND q) = ~~(p AND q) = ~(~p OR q)

    [This is similar to saying (a + b) = (-1)*(-1)*(a + b) = (-1)*(-a + -b)]

    I'll use N(q) for the proposition "Necessarily, q is true", which is equivalent to "q is true in every possible world". I'll use P(q) for the proposition "Possibly, q is true", which is equivalent to "There exists at least one possible world in which q is true."

    PROPERTIES OF N(x) AND P(x)
    N(x) = ~P(~x)

    Proof: If x is true in every possible world, then there can't be a world where it is false. Thus there does not exist a possible world where ~x holds. This is the definition of ~P(~x).

    Conversely, suppose that there are no possible worlds in which ~x holds. Then there are no possible worlds in which x is false. Thus x is true in every possible world. This is simply N(x)


    P(x) = ~N(~x)

    Suppose there is a possible world in which x is false; call it W. If N(~x) were true, ~x would hold in every possible world. In particular, ~x would hold in W. But x holds in W; this is a contradiction. Hence N(~x) must be false, thus ~N(~x) holds.

    Conversely, Suppose ~N(~x) holds. Then it is false that ~x holds in every possible world.



    DeMorgan's Modal Laws:
    ~(N(p) AND N(q)) = ~N(p) OR ~N(q)

    Suppose "N(p) AND N(q)" is false. That is, "p is true in every possible world AND q is true in every possible world" is false.

    Case 1: p is true in every possible world, but q isn't. Then N(q) is false. Hence ~N(q) is true. Thus ~N(p) OR ~N(q) is true.

    Case 2: p is true in every possible world, and so is q. Thus N(p) AND N(q). This contradicts our first assumption.

    Case 3: p isn't true in every possible world, but q is. Thus ~N(p). Hence ~N(p) OR ~N(q) is true.

    Case 4: p isn't true in every possible world, and neither is q. Thus ~N(p) (and also ~N(q)). Hence ~N(p) OR ~N(q) is true.


    In any case not entailing contradiction, ~N(p) OR ~N(q) is true.


    Conversely, suppose ~N(p) OR ~N(q).
    Case 1: ~N(p). Thus N(p) is false. Suppose "N(p) AND N(q)" is true; then N(p) is true. Contradiction. Thus ~(N(p) AND N(q)) holds.

    Case 2: ~N(q). Thus N(q) is false. By similar argument above, ~(N(p) AND N(q)) holds.

    In either case, ~(N(p) AND N(q)) holds.

    QED.


    LEMMA 1
    ~(P(x) AND P(y)) = ~P(x) OR ~P(y)

    Proof:
    ~(P(x) AND P(y))
    = ~(~N(~x) AND ~N(~y)) [Since P(x) = ~N(~x)]
    = N(~x) AND N(~y) [By DeMorgan's modal law]
    = ~P(x) AND ~P(y)
    -> ~P(x) OR ~P(y)

    Conversely,
    ~P(x) OR ~P(y)
    = N(~x) OR N(~y)

    Case 1: N(~x). Then ~x holds in every possible world. Now suppose "P(x) AND P(y)" is true. Then there is a possible world in which x is true; this contradicts N(~x). Hence "P(x) AND P(y)" must be false, i.e. ~(P(x) AND P(y)) holds.

    Case 2: N(~y). By similar argument above, ~(P(x) AND P(y)) holds.

    In either case, ~(P(x) AND P(y)) holds.

    QED


    LEMMA 2
    N(p AND q) -> N(p) AND N(q)

    Proof: Suppose N(p AND q). Then "p AND q" is true in every possible world. In particular, p is true in every possible world. Thus N(p) holds. By similar argument, N(q) holds. By definition of AND, N(p) AND N(q) holds.

    LEMMA 3
    P(p AND q) => P(p) AND P(q)

    Proof: Suppose P(p AND q). Then there exists at least one possible world where "p AND q" holds. Thus there exists at least one possible world where p holds, hence P(p) is true. By similar argument, P(q) holds. By definition of AND, P(p) AND P(q) holds.

    THE MAIN QUESTION
    Is the following true?

    N(p OR q) => N(p) OR N(q)


    It would seem not; just because "p OR q" must be true in every possible world shouldn't mean that p is true in every possible world, or q is true in every possible world. For example, take p = "There exists at least one human", and q = "There does not exist any humans". Then in any possible world, one of {p,q} will be true; thus, N(p OR q).

    But clearly N(q) is false, since the actual world has humans and is possible. And N(p) isn't obviously true, since we can imagine worlds where humanity never evolved.

    THE RUB

    N(p OR q)
    = N(~(~p AND ~q) [By reverse DeMorgan's law]
    = ~P~(~(~p AND ~q) [Since N(x) = ~P(~x)]
    = ~P(~p AND ~q) [since ~~x = x]
    -> ~(P(~p) AND P(~q)) [By lemma 3]
    -> ~P(~p) OR ~P(~q) [By lemma 1]
    = N(p) OR N(q) [since ~P(~x) = N(x)]

    QED


    This has strange implications. For instance, suppose p is a proposition true in the actual world. Consider "p OR ~p". This must be true in every possible world; hence N(p OR ~p). Then by the above argument, N(p) OR N(~p). If N(~p) were true, then p would be false in every possible world, including ours--a contradiction. Hence N(p) must be true.

    That is, anything true in our world is true in any possible world.
    If I am capable of grasping God objectively, I do not believe, but precisely because I cannot do this I must believe. - Soren Kierkegaard
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  2. #2
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    Re: Modal logic / Possible world hijinx?

    (In Lemma 1) How do you get from: ~(~N(~x) AND ~N(~y)) to N(~x) AND N(~y) by way of DeMorgan's modal law: ~(N(p) AND N(q)) = ~N(p) OR ~N(q)?

  3. #3
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    Re: Modal logic / Possible world hijinx?

    Quote Originally Posted by Galendir View Post
    (In Lemma 1) How do you get from: ~(~N(~x) AND ~N(~y)) to N(~x) AND N(~y) by way of DeMorgan's modal law: ~(N(p) AND N(q)) = ~N(p) OR ~N(q)?
    Starting with: ~(~N~x AND ~N~y)
    Apply DeMorgan's by negating each conjunct and taking their disjunction: ~~N~x OR ~~N~y
    Simplifying: N~x OR N~y


    EDIT: Oh, right. The proof in lemma 1 is wrong, but the conclusion is true, since N~x OR N~y = ~Px OR ~Py
    If I am capable of grasping God objectively, I do not believe, but precisely because I cannot do this I must believe. - Soren Kierkegaard
    **** you, I won't do what you tell me

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  4. #4
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    Re: Modal logic / Possible world hijinx?

    I think DeMorgan's law should be stated more succintly as:

    N~(X and Y) =>
    P(~X) or P(~Y)

    The second to last step of your clincher should actually be P(~~X) or P(~~Y)
    Because of the way you stated it it allowed for ambiguity. It allows you to abuse differences between ~P(X) and P(~X)
    Last edited by Soren; April 4th, 2012 at 11:07 PM.

  5. #5
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    Re: Modal logic / Possible world hijinx?

    Quote Originally Posted by Soren View Post
    I think DeMorgan's law should be stated more succintly as:

    N(X and Y) =>
    P(~X) or P(~Y)

    The N of my original statement (as opposed to yours) can be replaced with a ~P, which was what your original intention was, but because of the way you stated it it allowed for ambiguity.

    When it is stated this way your proof breaks down.
    How does your statement follow from DeMorgan's? It seems like you're reasoning this way:

    N(X and Y)
    = ~P~(X and Y) [N = ~P~]
    = ~P(~X or ~Y) [standard DeMorgan's on propositions]
    =? P(~X) or P(~Y) [?]
    If I am capable of grasping God objectively, I do not believe, but precisely because I cannot do this I must believe. - Soren Kierkegaard
    **** you, I won't do what you tell me

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  6. #6
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    Re: Modal logic / Possible world hijinx?

    Quote Originally Posted by CliveStaples View Post
    How does your statement follow from DeMorgan's? It seems like you're reasoning this way:

    N(X and Y)
    = ~P~(X and Y) [N = ~P~]
    = ~P(~X or ~Y) [standard DeMorgan's on propositions]
    =? P(~X) or P(~Y) [?]
    I'm sorry that was a typo (I have no access to a computer right now), but I don't think the way you used Lemma 3 was valid. The only reason that it is true is because when you "break it up" (for lack of a better term) it is referring to the same possible world. If you used subscripts for the Ps after you applied Lemma 3 (to know which world you are talking about) then you would end up with:
    It is necessary that p is true in world one or that q is in world one.

    The flaw in the proof arises from the fact that Lemma 3 isn't true when negated. Try using subscripts for the Ps with an "x" meaning "some unknown possible world", a number to mean one specific world, and a "!" to mean all worlds.

  7. #7
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    Re: Modal logic / Possible world hijinx?

    Quote Originally Posted by Soren View Post
    I'm sorry that was a typo (I have no access to a computer right now), but I don't think the way you used Lemma 3 was valid. The only reason that it is true is because when you "break it up" (for lack of a better term) it is referring to the same possible world. If you used subscripts for the Ps after you applied Lemma 3 (to know which world you are talking about) then you would end up with:
    It is necessary that p is true in world one or that q is in world one.

    The flaw in the proof arises from the fact that Lemma 3 isn't true when negated. Try using subscripts for the Ps with an "x" meaning "some unknown possible world", a number to mean one specific world, and a "!" to mean all worlds.
    You're very close. Lemma 3 isn't the problem; it's perfectly valid. The problem is that I used the classic fallacy of denying the antecendent; just because p -> q doesn't mean ~p -> ~q.

    Or here, P(p AND q) => P(p) AND P(q) doesn't mean ~P(~p AND ~q) -> ~(P(~p) AND P(~q)).
    If I am capable of grasping God objectively, I do not believe, but precisely because I cannot do this I must believe. - Soren Kierkegaard
    **** you, I won't do what you tell me

    HOLY CRAP MY BLOG IS AWESOME

  8. #8
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    Re: Modal logic / Possible world hijinx?

    Quote Originally Posted by CliveStaples View Post
    You're very close. Lemma 3 isn't the problem; it's perfectly valid. The problem is that I used the classic fallacy of denying the antecendent; just because p -> q doesn't mean ~p -> ~q.

    Or here, P(p AND q) => P(p) AND P(q) doesn't mean ~P(~p AND ~q) -> ~(P(~p) AND P(~q)).
    Yeah. It's hard to not get stuck in the mode of thinking it is like a math proof - but the rules are slightly different. I was thinking that ~P could just be replaced with something else (~p and ~q could be replaced with t and z, respectively, for example) like D, but it's a classifier, not a variable.

    P.S. I still think using subscripts would allow for more precision.

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    Re: Modal logic / Possible world hijinx?

    Quote Originally Posted by Soren View Post
    Yeah. It's hard to not get stuck in the mode of thinking it is like a math proof - but the rules are slightly different. I was thinking that ~P could just be replaced with something else (~p and ~q could be replaced with t and z, respectively, for example) like D, but it's a classifier, not a variable.

    P.S. I still think using subscripts would allow for more precision.
    I don't think I understand what you want the subscripts to mean. Are the modal operators subscripted? Do the subscripts refer to particular worlds, or particular propositions?

    And it is very much like a math proof (basic set theory is equivalent to boolean algebra). This is like forgetting to flip the inequality sign when you multiply both sides by a negative.
    If I am capable of grasping God objectively, I do not believe, but precisely because I cannot do this I must believe. - Soren Kierkegaard
    **** you, I won't do what you tell me

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    Re: Modal logic / Possible world hijinx?

    Quote Originally Posted by CliveStaples View Post
    I don't think I understand what you want the subscripts to mean. Are the modal operators subscripted? Do the subscripts refer to particular worlds, or particular propositions?
    The subscripts would refer to specific possible worlds.

    The truth of Lemma 3 is better represented if it would be written like this:
    Px(p AND q) => Px(p) AND Px(q)
    (With "x" referring to the same possible world)
    And:
    Px(~p AND ~q) => Px(~p) AND Px(~q)

    And (denying the antecedent of Lemma 3)
    ~P(p AND q) => ~Px(p) AND ~Px(q)
    Because
    ~Px(p) AND ~Px(q)
    Is the same as saying: It is false that in possible world x p is true AND in possible world x q is true.

    The following would then be true:
    ~Px(p) <=> Px(~p)
    N(p) <=> PALL(p)
    N(~p) => Px(~p)
    N(~p) <=> ~N(p) <=> ~P(p) => P(~p)

    When we use this subscript notation (it is easier to see where) your proof breaks down at the following (the bolded is where the meaning changes):

    N(p OR q)
    = N(~(~p AND ~q) [By reverse DeMorgan's law]
    = ~P~(~(~p AND ~q) [Since N(x) = ~P(~x)]
    = ~P(~p AND ~q) [since ~~x = x]
    -> ~(Px(~p) AND Px(~q)) [By lemma 3]
    -> ~Px(~p) OR ~Px(~q) [By lemma 1]
    => Px(~~p) OR Px(~~q) [Bolded red law above]
    => Px(p) OR Px(q)


    And the last statement self evidently follows from what you started with.

    And it is very much like a math proof (basic set theory is equivalent to boolean algebra). This is like forgetting to flip the inequality sign when you multiply both sides by a negative.
    I guess I just meant that it can be hard to catch denying the antecedent because I am used to doing math equations where the negative is interchangeable (because it is an equation) on both sides.
    Last edited by Soren; April 6th, 2012 at 07:36 PM.

  11. #11
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    Re: Modal logic / Possible world hijinx?

    Quote Originally Posted by Soren View Post
    The subscripts would refer to specific possible worlds.

    The truth of Lemma 3 is better represented if it would be written like this:
    Px(p AND q) => Px(p) AND Px(q)
    (With "x" referring to the same possible world)
    And:
    Px(~p AND ~q) => Px(~p) AND Px(~q)

    And (denying the antecedent of Lemma 3)
    ~P(p AND q) => ~Px(p) AND ~Px(q)
    Because
    ~Px(p) AND ~Px(q)
    Is the same as saying: It is false that in possible world x p is true AND in possible world x q is true.

    The following would then be true:
    ~Px(p) <=> Px(~p)
    N(p) <=> PALL(p)
    N(~p) => Px(~p)
    N(~p) <=> ~N(p) <=> ~P(p) => P(~p)

    When we use this subscript notation (it is easier to see where) your proof breaks down at the following (the bolded is where the meaning changes):

    N(p OR q)
    = N(~(~p AND ~q) [By reverse DeMorgan's law]
    = ~P~(~(~p AND ~q) [Since N(x) = ~P(~x)]
    = ~P(~p AND ~q) [since ~~x = x]
    -> ~(Px(~p) AND Px(~q)) [By lemma 3]
    -> ~Px(~p) OR ~Px(~q) [By lemma 1]
    => Px(~~p) OR Px(~~q) [Bolded red law above]
    => Px(p) OR Px(q)


    And the last statement self evidently follows from what you started with.
    This doesn't seem like useful notation. It seems like you want Px(p AND q) to mean "p AND q is true in the possible world x". But you might as well just say X(p AND q) to mean "p AND q is true in world X".

    You don't have to keep naming the worlds in which these propositions are true; it's not useful information for the most part. All that matters is the existence. Saying "if p is possibly true, it is not necessarily false" is the same as saying "If there is a possible world X where p is true, p is not necessarily false". You don't gain anything with your notation.

    I guess I just meant that it can be hard to catch denying the antecedent because I am used to doing math equations where the negative is interchangeable (because it is an equation) on both sides.
    Except that implication is equivalent to subset, which is equivalent to <=. So it works pretty consistently with the operations you can do on the <= relation.
    If I am capable of grasping God objectively, I do not believe, but precisely because I cannot do this I must believe. - Soren Kierkegaard
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    Re: Modal logic / Possible world hijinx?

    Quote Originally Posted by CliveStaples View Post
    This doesn't seem like useful notation. It seems like you want Px(p AND q) to mean "p AND q is true in the possible world x". But you might as well just say X(p AND q) to mean "p AND q is true in world X".
    Well the P is there to signify that it represents one Possible world. According to your logic we might as well do away with Ps all together, and assume that any variable is being described in a possible world unless stated otherwise - because there is less writing involved.
    You don't have to keep naming the worlds in which these propositions are true; it's not useful information for the most part. All that matters is the existence.
    What makes you think that all that mattera is the existence? The purpose of modal logic is, more or less, to gain a better sense and understanding of the implications of possiblility and necessity, and if a notation was used were it demonstrated more of what was going on EXACTLY - then it would be more fulflling to its purpose, and therefore more useful.
    Saying "if p is possibly true, it is not necessarily false" is the same as saying "If there is a possible world X where p is true, p is not necessarily false". You don't gain anything with your notation.
    Well I can see why it is redundant in the example you gave, and I would agree that in that paricular instance it isnt any more useful. But, I showed how it has more "provability" power (in my system the converse of Lemma 3 has implication, but in yours it doesn't) - and I also gave an example, which was your proof, of how my system was better.

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    Re: Modal logic / Possible world hijinx?

    Quote Originally Posted by Soren View Post
    Well the P is there to signify that it represents one Possible world. According to your logic we might as well do away with Ps all together, and assume that any variable is being described in a possible world unless stated otherwise - because there is less writing involved.
    P(x) means "There exists a possible world where x is true." Using your notation to represent one of the possible worlds in which x is true doesn't seem useful unless you want to know other things about x in particular.

    What makes you think that all that mattera is the existence? The purpose of modal logic is, more or less, to gain a better sense and understanding of the implications of possiblility and necessity, and if a notation was used were it demonstrated more of what was going on EXACTLY - then it would be more fulflling to its purpose, and therefore more useful.
    Right, if we were interested in describing a particular possible world, your notation would be useful. But what we're interested in are statements like "P(x and y) <-> P(x) and P(y)". If we used your notation, we'd need arbitrary subscripts on P(x) and P(y), since we don't know if they're true in the same possible world or not.

    "There exists a possible world where "x and y" holds <-> There exists a possible world where x holds and there exists a possible world where y holds" is what the statement really means. You run into specification problems with your notation, since P-sub-x might equal P-sub-y. So just because you tack on a subscript doesn't mean that you're actually specifying distinct possible worlds. This is an unnecessary complication when all we care about is the existence, as in the proof above. That is, all we care about is that there's a world where x and y are true, there's a world where x is true, and there's a world where y is true.

    Well I can see why it is redundant in the example you gave, and I would agree that in that paricular instance it isnt any more useful. But, I showed how it has more "provability" power (in my system the converse of Lemma 3 has implication, but in yours it doesn't) - and I also gave an example, which was your proof, of how my system was better.
    But it only helps in one direction--saying that P-sub-x(a AND b) implies P-sub-x(a) and P-sub-x(b). It's good when you're doing specification--that is, "There's a world where a and b are true. In particular, in this world, a is true." It doesn't help when you're trying to show that the truth of a and b in one possible world implies something about a distinct possible world.

    If we used your notation consistently, we wouldn't be able to just say "There exists a possible world where x is true." We'd have to specify it each time, and we wouldn't know when worlds were intersecting--maybe x and y are truths such that every world in which x holds, y also holds, but there are worlds where y holds and x doesn't. Every possibility statement that isn't assumed to be about the same possible world has to have a different subscript attached to it--but we wouldn't know that they were actually distinct.

    There's a reason why we use "There exists" and "For all" in mathematics. There's a reason we don't use a subscript on those backward E's. Your notation would get rid of all simple "There exists" statements regarding possible world, without adding useful information. The only thing your notation does is change "Suppose P(x); then there exists a possible world where x is true, call it W; [some statement relating to x or W]" to "Suppose P-sub-w(x). [Some statement relating to x or W]." But it's only advantageous when you're making a statement about W. If you're only making a statement about x, or about some other possible world V, then knowing that x is true specifically in W is irrelevant.
    If I am capable of grasping God objectively, I do not believe, but precisely because I cannot do this I must believe. - Soren Kierkegaard
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  14. #14
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    Re: Modal logic / Possible world hijinx?

    I must say, I had a lot of fun with this; it's a fantastic improvement over those 1=2 "proofs" that pop up from time-to-time.

    Quote Originally Posted by CliveStaples
    LEMMA 1
    ~(P(x) AND P(y)) = ~P(x) OR ~P(y)

    Proof:
    ~(P(x) AND P(y))
    = ~(~N(~x) AND ~N(~y)) [Since P(x) = ~N(~x)]
    = N(~x) AND N(~y) [By DeMorgan's modal law]
    Here, it seems like the next step should have been N(~x) OR N(~y).

    = ~P(x) AND ~P(y)
    -> ~P(x) OR ~P(y)
    And there goes my hopes of finding the flaw in this proof on the first lemma. It seems that you meant to type "OR" instead of "AND" anyway.

    Lemma 2 looks fine. So does lemma 3. As does your reasoning suggesting that N(p OR q) <=> N(p) OR N(q).

    As for the rub, I had to write this out line-by-line on a piece of paper, scrawling Venn diagrams and notes. The problem I found was with this line:

    Quote Originally Posted by CliveStaples
    N(p OR q)
    = N(~(~p AND ~q) [By reverse DeMorgan's law]
    = ~P~(~(~p AND ~q) [Since N(x) = ~P(~x)]
    = ~P(~p AND ~q) [since ~~x = x]
    -> ~(P(~p) AND P(~q)) [By lemma 3]
    -> ~P(~p) OR ~P(~q) [By lemma 1]
    = N(p) OR N(q) [since ~P(~x) = N(x)]
    Lemma 3 says that:

    P(x AND y) -> P(x) AND P(y).

    Fair enough.

    What the rub requires is that:

    ~P(x AND y) -> ~(P(x) AND P(y)).

    [If I use ~p and ~q in place of x and y for this example, my head may very well explode.]

    However, in order to show that, you'd need something stronger than Lemma 3. Lemma 3 proves the sufficiency of P(x AND y) for (P(x) AND P(y)), but doesn't prove its necessity. Without establishing a logical biconditional, you can't say the negation of the first implies the negation of the second.

    In fact, it can be reasoned intuitively why such a biconditional couldn't be established:

    There might not exist a world where both of {x,y} are true, but it doesn't follow that there is not a world where x is true and a separate world where y is true.

    For instance, take:
    x = Humans exist
    y = ~x

    ~P(x AND y) is definitely true (that there is no possibility for a world to contain a contradiction). However, ~(P(x) AND P(y)) is not obviously true; we know P(x) is true from our own world, but it is conceivable on another world that humans never evolved, like you pointed out, so P(y) may also be true.

    I must say, you almost had me thinking that you'd found a way to confound the many worlds interpretation of quantum mechanics. By the way, is a mathematical conception of multiple worlds a one-off invention for this thread, or a field in which you've done more work? If it's the latter, I'd love to see more.

  15. #15
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    Re: Modal logic / Possible world hijinx?

    Quote Originally Posted by LL66 View Post
    I must say, I had a lot of fun with this; it's a fantastic improvement over those 1=2 "proofs" that pop up from time-to-time.


    Here, it seems like the next step should have been N(~x) OR N(~y).


    And there goes my hopes of finding the flaw in this proof on the first lemma. It seems that you meant to type "OR" instead of "AND" anyway.

    Lemma 2 looks fine. So does lemma 3. As does your reasoning suggesting that N(p OR q) <=> N(p) OR N(q).

    As for the rub, I had to write this out line-by-line on a piece of paper, scrawling Venn diagrams and notes. The problem I found was with this line:



    Lemma 3 says that:

    P(x AND y) -> P(x) AND P(y).

    Fair enough.

    What the rub requires is that:

    ~P(x AND y) -> ~(P(x) AND P(y)).

    [If I use ~p and ~q in place of x and y for this example, my head may very well explode.]

    However, in order to show that, you'd need something stronger than Lemma 3. Lemma 3 proves the sufficiency of P(x AND y) for (P(x) AND P(y)), but doesn't prove its necessity. Without establishing a logical biconditional, you can't say the negation of the first implies the negation of the second.

    In fact, it can be reasoned intuitively why such a biconditional couldn't be established:

    There might not exist a world where both of {x,y} are true, but it doesn't follow that there is not a world where x is true and a separate world where y is true.

    For instance, take:
    x = Humans exist
    y = ~x

    ~P(x AND y) is definitely true (that there is no possibility for a world to contain a contradiction). However, ~(P(x) AND P(y)) is not obviously true; we know P(x) is true from our own world, but it is conceivable on another world that humans never evolved, like you pointed out, so P(y) may also be true.

    I must say, you almost had me thinking that you'd found a way to confound the many worlds interpretation of quantum mechanics. By the way, is a mathematical conception of multiple worlds a one-off invention for this thread, or a field in which you've done more work? If it's the
    latter, I'd love to see more.
    Well the fallacy is the same as saying "If an animal is a Pitbull, then it is a dog" but this doesn't mean that if an animal isn't a pitbull, then it can't be a dog. (Formally it is denying the antecedent). This formalism of possible worlds is known as modal logic, and I believe Clive gave an informative link on this subject in the beginning of his OP.

  16. #16
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    Re: Modal logic / Possible world hijinx?

    Quote Originally Posted by Soren View Post
    Well the fallacy is the same as saying "If an animal is a Pitbull, then it is a dog" but this doesn't mean that if an animal isn't a pitbull, then it can't be a dog. (Formally it is denying the antecedent).
    Yes. Aside from my failure to name the fallacy, I think we're making the same point.

    (As an aside, I've taken the meaning of this forum to be the presentation of challenges to which the OP already knows the answer, so I didn't read any replies until after I posted, at which point I realised that there was a more succinct way to make my point. If the thread was just supposed to stop once the answer was reached, then I apologise for being redundant.)

    Quote Originally Posted by Soren
    This formalism of possible worlds is known as modal logic, and I believe Clive gave an informative link on this subject in the beginning of his OP.
    Thanks! Somehow, I completely missed that link.

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    Re: Modal logic / Possible world hijinx?

    Quote Originally Posted by LL66 View Post
    Yes. Aside from my failure to name the fallacy, I think we're making the same point.

    (As an aside, I've taken the meaning of this forum to be the presentation of challenges to which the OP already knows the answer, so I didn't read any replies until after I posted, at which point I realised that there was a more succinct way to make my point. If the thread was just supposed to stop once the answer was reached, then I apologise for being redundant.)
    Lol, no worries. When I first read this thread I couldn't tell at first whether or not this was something that Clive stumbled upon and didn't know the answer to, or if it was a riddle that he knew the error of. Generally you are correct and it is the latter (and the solution is written multiple times); I wasn't trying to insinuate that you were being redundant or anything; I was just trying being helpful.

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    Re: Modal logic / Possible world hijinx?

    This is an interesting subject.

    As I was reading the first post and trying to grasp the concept, I stopped at this line, and I am hoping you can clarify it for me:

    (p AND q) = ~~(p AND q) = ~(~p OR q)

    Perhaps I am just looking at this from an algebraic point of view. I understand the first one (p AND q), and I understand how it is equal to ~~(p AND q). I cannot determine how either of those is equal to ~(~p OR q). The example you use for comparison is:

    [This is similar to saying (a + b) = (-1)*(-1)*(a + b) = (-1)*(-a + -b)]

    Was this just a typo? Did you mean to put ~(~p OR ~q)? I don't mean to pick at it, but this point is preventing me from moving on to the next bit in your post.
    It is not our abilities in life that show who we truly are; it is our choices. Albus Dumbledore in Harry Potter and the Chamber of Secrets

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    Re: Modal logic / Possible world hijinx?

    Quote Originally Posted by ladykrimson View Post
    This is an interesting subject.

    As I was reading the first post and trying to grasp the concept, I stopped at this line, and I am hoping you can clarify it for me:

    (p AND q) = ~~(p AND q) = ~(~p OR q)

    Perhaps I am just looking at this from an algebraic point of view. I understand the first one (p AND q), and I understand how it is equal to ~~(p AND q). I cannot determine how either of those is equal to ~(~p OR q). The example you use for comparison is:

    [This is similar to saying (a + b) = (-1)*(-1)*(a + b) = (-1)*(-a + -b)]

    Was this just a typo? Did you mean to put ~(~p OR ~q)? I don't mean to pick at it, but this point is preventing me from moving on to the next bit in your post.
    Right, so under DeMorgan's law, we know how to 'distribute' a negation over an OR or an AND:

    ~(p AND q) = ~p OR ~q
    ~(p OR q) = ~p AND ~q

    e.g., If "I'm tall AND blonde" is false, either I'm not tall, or I'm not blonde; if "He's hiding here OR he's hiding there" is false, then he isn't hiding here, and he isn't hiding there.

    So if we were looking at ~~(p AND q), we would 'distribute' the first ~ through, viewing the expression as ~(~(p AND q)), and get:

    ~(~p OR ~q)
    If I am capable of grasping God objectively, I do not believe, but precisely because I cannot do this I must believe. - Soren Kierkegaard
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    Re: Modal logic / Possible world hijinx?

    Sorry, I meant to post on this ages ago. The flaw in this is an ordering ambiguity, so to speak.

    1.) N(p OR q)
    2.) = N(~(~p AND ~q) [By reverse DeMorgan's law]
    3.) = ~P~(~(~p AND ~q) [Since N(x) = ~P(~x)]
    4.) = ~P(~p AND ~q) [since ~~x = x]
    5.) -> ~(P(~p) AND P(~q)) [By lemma 3]
    6.) -> ~P(~p) OR ~P(~q) [By lemma 1]
    7.) = N(p) OR N(q) [since ~P(~x) = N(x)]

    I remember when I first looked at this a few months ago, I just couldn't see the problem. I thought I responded saying the answer is somewhere in the middle steps, but apparently I never did. Upon finding this out, I took the 30 minutes to finally spot it. My conclusion is that ultimately you've commit an inverse fallacy from step 4 to step 5.

    If we write this out properly, we can see the error:

    ~P(~p AND ~q) -> ~(P(~p) AND P(~q))

    But this is simply false. The converse of this statement is true, but you're ignoring the existence of the ~ in front of both terms, and this switches the ordering by the "->". Note that the following statement is true:

    P(~p AND ~q) -> (P(~p) AND P(~q))

    But if we have two negation symbols on both sides, this reverses it to:

    "P(~p AND ~q) => (P(~p) AND P(~q))" equals "~(P(~p) AND P(~q)) => ~P(~p AND ~q)" [This is modus tollens, or "A => B" equals "~B => ~A"]

    So the error was assuming you could just substitute in what you're calling "Lemma 3", but you can only do that if Lemma 3 were about logical equivalency, not logical implication. The implication binary operator has ordering. Unfortunately, again, the negation signs act to obstruct this and reverse the ordering --and for good reason. We see that by switching the order, as required, it instead derives a true relation, if we follow through correctly on the remaining steps:

    1.) N(p OR q)
    2.) = N(~(~p AND ~q) [By reverse DeMorgan's law]
    3.) = ~P~(~(~p AND ~q) [Since N(x) = ~P(~x)]
    4.) = ~P(~p AND ~q) [since ~~x = x]
    5.) ~(P(~p) AND P(~q)) => ~P(~p AND ~q) [By lemma 3 & applying modus tollens]
    6.) (~P(~p) OR ~P(~q)) => N~(~p AND ~q) [By lemma 1, on LHS, "~P = N~" on RHS]
    6a.) (~P(~p) OR ~P(~q)) => N(p OR q) [Another De Morgan dual on RHS]
    7.) N(p) OR N(q) => N~(~p OR ~q) [since ~P(~x) = N(x)]

    Which has the chief virtue of being true. Q.E.D.
    "Those who can make you believe absurdities, can make you commit atrocities." --Voltaire

 

 
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