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Math Help - valid sequent

  1. #1
    Newbie
    Joined
    Jan 2009
    Posts
    15

    valid sequent

    Hi,

    I have used the semantic tableau procedure to check if the following sequent is valid. I get the answer that it is valid, however I am not sure if it is indeed valid and I havent made an error while writing the tableau. I havent included the tableau in this post as I don't know how to include it but the sequent is the following:

    <br />
(\exists x A(x)) \rightarrow B \vdash \forall x (A(x) \rightarrow B)<br />

    Is this sequent indeed valid?

    Another sequent:

    <br />
(\forall x A(x)) \rightarrow B \vdash \exists x (A(x) \rightarrow B)<br />

    Also for this 2nd sequent I get the result that it is valid.

    Am I correct for both sequents?


    Thanks for any help.
    Last edited by sanv; February 25th 2009 at 12:59 PM.
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  2. #2
    Senior Member
    Joined
    Nov 2008
    Posts
    394
    Quote Originally Posted by sanv View Post
    <br />
(\exists x A(x)) \rightarrow B \vdash \forall x (A(x) \rightarrow B)<br />
    Assume x does not occur free in B.

    By the deduction and generalization theorems,

    <br />
\exists x A \rightarrow B \vdash \forall x (A \rightarrow B),<br />
    <br />
\{(\exists x A \rightarrow B), A\} \vdash B.<br />

    Since A \vdash \exists x A (using a contraposition and verify this), the above formula is valid.


    <br />
(\forall x A(x)) \rightarrow B \vdash \exists x (A(x) \rightarrow B)<br />
    1. <br />
\forall x A \rightarrow B \vdash \exists x (A \rightarrow B),<br />
    2. <br />
\neg \exists x \neg A \rightarrow B \vdash \exists x (A \rightarrow B),<br />
    3. <br />
\neg \exists x \neg A \rightarrow B \vdash A \rightarrow B,<br />
    4. <br />
\{(\neg \exists x \neg A \rightarrow B), A\} \vdash B, (\text{by deduction theorem})<br />

    Since  A \vdash \neg \exists x \neg A (verify this), the above formula is valid.
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