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There Exists Only One

  1. Jul 2, 2013 #1
    1. The problem statement, all variables and given/known data
    Express each of these system specifications using predicates, quantifiers, and logical connectives, if necessary.

    a) Every user has access to exactly one mailbox.


    2. Relevant equations



    3. The attempt at a solution

    It is typical of my book to not answer questions as given with the unique existential quantifier [itex]\exists ![/itex]. For instance, the answer to the question above is [itex]∀u∃m(A(u, m)∧∀n(n \ne m→¬A(u, n)))[/itex]. However, I am not convinced that this form assures that only one m exists for every u. Isn't it still possible that [itex]m_0[/itex] and[itex]m_1[/itex] are two elements in the domain of the variable that make the statement, implying that there doesn't exists one and only one value of m for every u?
     
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  3. Jul 2, 2013 #2

    D H

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    No. If those m0 and m1 are distinct (i.e., m0 ≠ m1), then both of them cannot satisfy A(u,m1) per the second part of the condition, [itex]\forall n(n\ne m \rightarrow \neg A(u,n))[/itex].
     
  4. Jul 2, 2013 #3
    Well, why couldn't every n correspond to m0, and then every n also correspond to m1?
     
  5. Jul 2, 2013 #4

    verty

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    It ranges over EVERYTHING, everything in the universe of discourse (or at least, everything that it can represent).
     
  6. Jul 8, 2013 #5
    Verty, I am not certain how that aids in answering my question.
     
  7. Jul 8, 2013 #6

    haruspex

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    You are guaranteed the existence of m0, say, such that [itex]A(u, m_0)∧∀n(n \ne m_0→¬A(u, n)))[/itex]. Suppose m1 (≠m0) satisfies [itex]A(u, m_1)[/itex]. But we know [itex]∀n(n \ne m_0→¬A(u, n)))[/itex]. Since n can be m1, and [itex]A(u, m_1)[/itex], it follows that [itex]¬A(u, m_1)))[/itex].
     
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