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A statement that is true AND false?

  1. Jan 9, 2007 #1
    Given: my bedroom has no door.

    Statement 1: My bedroom door is open.
    Statement 2: My bedroom door is closed.

    [Definition: closed is equivalent to not open. So statement 2 is the negation of statement 1.]

    Now when I evaluate the truthfulness of statements 1 and 2...

    We can translate this to a more formal logical setting by introducing the unary predicate O standing for openness and constant d, which means my bedroom door.

    In this setting, statement 1 is Od and statement 2 is -Od [where - is a negation symbol]. I'm trying to figure out the truthfulness of the two statements Od and -Od.

    This is where I get stuck because I suspect that BOTH Od and -Od are vacuously true as the set of all possible states of the door, which would usually have two elements in it, open and not open, is empty.

    Since -Od is true, albeit vacuously, that means --Od is false. Since --Od is equivalent to Od, Od is false. And therefore, Od is true and false.

    What is the flaw with my reasoning? This smells to me like the "proof" that 1=2...
  2. jcsd
  3. Jan 9, 2007 #2


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    In logic, a "statement" is defined as a sentence that is EITHER true OR false. In the case that you have no door then the sentence "My door is open" is not a statement because it talks a non-existant object. Similarly, the sentence that "That unicorn is green" is not a statement.
    Last edited by a moderator: Jan 20, 2007
  4. Jan 9, 2007 #3
    If my room had a door, then Od suddenly becomes a statement?
  5. Jan 9, 2007 #4
    Check the law of the excluded middle.

    For instance: http://en.wikipedia.org/wiki/Law_of_excluded_middle" [Broken]
    Last edited by a moderator: May 2, 2017
  6. Jan 9, 2007 #5
    Um..I am aware of the law of the excluded middle. That's why this whole thing is a paradox.

    The question is what is the flaw with the reasoning that arrives at Od being true and false.

    More generally....

    Let's say S is a set. P is a unary predicate on S if P is any subset of S just as a relation or binary predicate is any subset of S^2.

    We write Px if x is an element of P just as if R is a relation we write xRy if (x,y) is in R.

    Now suppose P=Ø and x is any set. Actually now it would seem like I'd have to say that -Px is true as x is not in Ø. And I also must say that Px is false as x is not in Ø.

    Specifically, S is the set of all possible states of my door. If there is a door, then S has two elements and if there is no door, S=Ø. That forces O to also be empty, given that there is no door. OK, so O=Ø but since Ø is a subset of Ø, O is a genuine unary predicate on S, contrary to what HallsOfIvy is suggesting.

    By the more general paragraph above, I must now say that -Od is true and Od is false. This resolves the "paradox." Since -Od is true, --Od is false and --Od is equivalent to Od; so Od is false. There is no contradiction: Od is false but NOT true; so Od is NOT true and false.
  7. Jan 9, 2007 #6
    Then you ought to understand that your situation is not a paradox. :smile:

    If you have a statement and it is defined that that statement is either true or not true you imply the law of excluded middle.
    If you leave it out you can make millions of "paradoxes" because there is an alternative option.
  8. Jan 9, 2007 #7
    Right. It's not a paradox. That's why I used quotes and wrote "paradox" at the end. A true paradox has no resolution in my book.
  9. Jan 9, 2007 #8


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    It depends on precisely how you intend to formalize what you've written.

    The way I think I'm happiest is if you defined a language with the constant symbol "My bedroom door" and two relations "___ is open" and "___ is closed". Statement 1 and statement 2 are statements within that language. And presumably, you've implicity defined a theory that contains some axioms capturing just what it means for something to be your bedroom door.

    Now, your problem is that, since you don't have a bedroom door, there does not exist a model of your theory in reality. (at least, one that maps "My bedroom door" to something that would count as your bedroom door)

    Er, the law of noncontradiction is the very reason this is an apparent paradox: the law of noncontradiction contradicts the fact he has apparently proven both Od and -Od.

    (Noncontradiction is what you meant here -- having both Od and -Od true is perfectly consistent with the law of the excluded middle)

    Not so -- that "P or ~P" is true for one particular statement P doesn't mean it's true for every statement.
    Last edited: Jan 9, 2007
  10. Jan 9, 2007 #9
    How's it going, Hurkyl...long time to see, huh?

    Anyway, what is the problem with how I formalized it:
    The way I formalized it, there is no contradiction; so I satisfied at least myself. :wink: This seems to be a good example of abuse of the vacuous truth concept.

    In other words, I think I do have a mathematical model for the situation in which there is no door: this causes S to be empty whereas if there were two states, S would have two elements and then Od and -Od would both be contingent on whether the door is in reality open or closed. Without a door, it is discernible that Od is false. In English, this means "the door is open" is false...and it's false, in real terms, because there is no door.
  11. Jan 9, 2007 #10


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    A quick summary in English: you're using the predicate O as if it were

    "______ is an open bedroom door belonging to me"

    that operates on the class of all things, rather than the predicate

    "_____ is open"

    that operates on the class of your bedroom doors.

    Hrm, how'd I miss that other post?


    For this notation to make sense, it must be already known that x is an element of S, or that (x, y) is an element of S². Relations1 have the same syntactic rules as functions: an expression involving a relation symbol is only well-defined if the arguments lie in the domain of the relation.

    In particular, you can't say this: you defined P to be a unary relation on S, so Px is only well-defined if x is known to be of type S. That is, if it is a (possibly generalized) element of x.

    Now... if you instead defined P to be a unary relation on the class of all sets, then Px would make sense if x was an arbitrary set.

    And this is straight to the point: -Od and Od cannot be well-defined terms. In fact:

    d, all by itself, is not a well-defined term.

    Now, if you were trying to say "-O is the identically true predicate on S", then you'd be correct. Of course, you should be aware that all predicates on S are equal: -O is also the identically false predicate on S. O is also both identically false and identically true. This isn't a problem, because:

    (for all x in the S: false) = true

    1: I'm used to using the terms relation and predicate interchangably -- so there can be binary relations, ternary relations, unary relations, nullary relations, et cetera.
    Last edited: Jan 9, 2007
  12. Jan 9, 2007 #11


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    What does the symbol d denote? It has to denote something. It can't denote your bedroom door, because your bedroom has no door, so your bedroom door is not something. A little more formally, the description "your bedroom door" refers to nothing, thus your universe of discourse cannot have any element describable as "your bedroom door" and since d must refer to some object in your universe of discourse, d cannot stand for your bedroom door.
  13. Jan 10, 2007 #12
    That does seem to be the flaw with my reasoning. Thanks!
  14. Jan 15, 2007 #13
    Use terniary logic!

    It has 'truth' values, TRUE, FALSE and NULL

    NULL means simply: unknown or not applicable.

    Only thing you have to know is how the truth table looks for calculating truth values that contain NULL values.



  15. Jan 20, 2007 #14
  16. Feb 22, 2011 #15
    Its true if you want it to be true, its false if you want it to be false; its a vacuous statement :
    Its vacuous because you are assigning meaning to something that does not exist; or put another way, you are pretending that something that doesn't exists, does exists; therefore, everything that comes afterwards is vacuously true (or false, because the opposite (false) is true if you want it to be; i.e. if you want it to be false, then its true that its false, visa versa).
    Its really that simple.... forget about law of exclusion of middle.
  17. Feb 22, 2011 #16
    You are combining two statements each with their own truth value and asking questions about the validity of the combined statement, based upon the truth value of only one of the statements.

    Many paradoxes stem from this proceedure which is fraught with stumbling blocks.

    Consider something similar.

    "For any statement there is always the option 'not applicable'."

    Russell's paradox(es) also stems from this issue.

    As to those whose getout is that something does not exist - Logic should be tougher than that. It should be demonstrably applicable to a 'thought experiment'.
  18. Feb 22, 2011 #17
    Two mutually exclusive statements cannot both be true, but I'm not sure they can't both be false.
  19. Feb 23, 2011 #18


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    Well... Obviously it is possible that two mutually exclusive statements can be false at the same time. Consider "there are two apples in the basket" and "there are three apples in the basket". Mutually exclusiveness isn't the best analogy here.
  20. Feb 23, 2011 #19


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    If you follow Russell and Whitehead (Principia Mathematica), then
    "My bedroom door is open"
    is equivalent to
    "My bedroom door d exists" and "d is a member of the set of all open doors".

    If you have no door, the statements "My bedroom door is open" and "My bedroom door is closed" are both false, because "My bedroom door d exists" is false.

    Even if you define the predicate function "is closed" as the negation of the function "is open", the negation of "My bedroom door is open" (as intepreted by R&W) is NOT "My bedroom door is closed." Work out for yourself what is actually is, from the above definition.
  21. Feb 23, 2011 #20
    To qualify as a set of all open doors, surely this set must include all imaginary doors as well as real ones?
  22. Feb 23, 2011 #21
    I don't know about that. I keep all my imaginary doors closed and locked. You can't be too careful these days.
  23. Apr 5, 2011 #22
    Agreed, ~(My bedroom door is open) is true and ~(My bedroom door is closed) is true.
    There is no thing that (My bedroom door) is, if there is no door in my bedroom.
    We cannot say what My bedroom door is, but, we can say what it is not.

    Russell's analysis of 'The present king of France' makes this point clear.
    The present king of France is bald, and, The present king of France has hair on his head, are both false.

    See: On Denoting, B. Russell, 1905, ...it is online.

    (x is bald <-> ~(x has hair on his head)) iff (x exists).
    (x has hair on his head <-> ~(x is bald)) iff (x exists).

    That is to say..
    (My bedroom door is closed <-> ~(My bedroom door is open)) iff (My bedroom door exists).

    If there is no door in My bedroom, then My bedroom door does not exist.
    There is no proposition that has 'My bedroom door' as its subject that is true.

    G(the x:Fx) -> E!(the x:Fx), .. is a theorem of (FOPL=).
    :.~(E!(the x:Fx)) -> ~(G(the x:Fx)), for all G..is a theorem.

    ie. Ey(Ax(x=y <-> Fx) & Gy) -> EyAx(x=y <-> Fx), is a theorem.
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