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Math Help - Understanding Existential Instantiation

  1. #1
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    Understanding Existential Instantiation

    (x) (...x...)
    _____________
    ...a...
    Existential instantiation says, "For any sentence with variable x, constant symbol does not appear elsewhere in the knowledge base, we can replace x with constant symbol a".

    My question is
    1. why "a" should not appear in the knowledge base?
    2. What if we don't have any more symbol available, so we cannot pick up new symbol for existential quantifier?

    Thanks.
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  2. #2
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    Quote Originally Posted by crux View Post
    [LEFT]([IMG]My question is
    1. Why "a" should not appear in the knowledge base?
    2. What if we don't have any more symbol available, so we cannot pick up new symbol for existential quantifier?
    Lets begin with the second question.
    In fact we have an infinite collection of symbols.
    We can use one symbol with subscripts on it.

    For the first question, I have never seen existential instantiation put that way.
    I can only guess what the textbook means.
    Let me give you an example.
    There is a brown object.
    There is a dog.
    Therefore, there is a brown dog.
    Clearly that is not a valid argument.
    But without the rule for existential instantiation we could us the same constant twice and arrive at that conclusion as valid.
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  3. #3
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    1) Lets say you have a proof like:

    n. P(c)
    m. (∃x)(Fx)

    You have to instantiate to a different constant than c because who says that F(c) is true? All you know is that something in the domain falls under the predicate F. You can't assume that it's the element named by c.

    2) Your constant symbols are infinite so you won't run out.
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  4. #4
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    Quote Originally Posted by crux View Post
    (x) (...x...)




    _____________
    ...a...

    Existential instantiation says, "For any sentence with variable x, constant symbol does not appear elsewhere in the knowledge base, we can replace x with constant symbol a".


    My question is
    1. why "a" should not appear in the knowledge base?
    2. What if we don't have any more symbol available, so we cannot pick up new symbol for existential quantifier?


    Thanks.
    Perhaps the following examples will help to answer your problem ,although in a non axiomatic development of the predicate calculus there is a rule for existential elimination or instantiation:


    We know that the following statement is true:


    Everything has a mother.


    This can be expressed in the following formula:

    \forall y\exists x [ M(xy)] ,where M(xy) means x is the mother of y.


    Now since everything has a mother Alice must have a mother .Thus by dropping the Universal quantifier in the above formula we have:

    \exists x[ Mxa)], where M(xa) now means Alice has a mother.



    And now here is the crucial point of the whole case .If we put x=a in our existential elimination we will have as a result:


    ........................M(aa).....which means that Alice is her own mother ,not a possible case.


    Or even worst if we use Existential introduction on M(aa) we get the formula:



    ................... \exists x [ M(xx)], which means that:


    SOMETHING is its own mother. Not true.


    THE above derivation can be written in the following short way:



    \forall y\exists x [ M(xy)...........................................1


    \exists x [ M(xa)].........U.E ,1....................................2

    M(aa)...........................................H ( for E.E).............................3



    \exists x[ M(xx)]........E.I, 3......................................4


    \exists x [ M(xx)].........E.E, 2, 3-4.(Incorrect).............5




    Where U.E is for Universal Elimination , E.I is for Existential Introduction,E.E is for Existential Elimination and H for Hypothesis.


    In another example lets assume that:


    Someone is rich: \exists x [ R(x) ]............................1



    Assume also: Alice is poor : P(a).............................................. ....2


    If in assumption (1) we put x=a for Existential instantiation we get :



    R(a).............................................. .............................................3


    AND from (2) and ( 3) and by the law of addition introduction we have :



    P(a) & R(a).............................................. ....................................4


    and from (4) and by Existential Introduction we have:


    \exists x [ P(x) & R(x)] someone is rich and poor.Again incorrect


    Finally there are other couple of cases to cover
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