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Math Help - kleene-closure

  1. #1
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    kleene-closure

    Hello, I stuck with this exercise

    The kleene-closure of an Language L ist the smallest language, which contines L as subset and the empty word as an element and which is closend under concatenation. I have to profe by solving a-e

    a) L \subseteq L*
    b) L* \cdot L* \subseteq L*
    c) For every L' (over same Alphabet like L) with L \subseteq L' , \epsilon \in L' and L' \cdot L' \subseteq L' obtains L* \subseteq L'.

    I think I solved a) by showing

    L = {w1 | w1 \in L} L*={w1,.....,wn | n \in N, w_i \in L for all i = 1....n}

    => w1 \in L and w1 \inL* => L \subseteqL*

    b) and c) I stuck and any help will be appreciated


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  2. #2
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    Re: kleene-closure

    Quote Originally Posted by suslik View Post
    b) L* \cdot L* \subseteq L*
    Suppose w\in L^*\cdot L^*. Then w=uv for some u, v\in L^*, i.e., u and v are concatenations of words from L. What can you way about uv?

    Quote Originally Posted by suslik View Post
    c) For every L' (over same Alphabet like L) with L \subseteq L' , \epsilon \in L' and L' \cdot L' \subseteq L' obtains L* \subseteq L'.
    The definition of L^* says that it is the smallest language satisfying these properties.
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  3. #3
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    Re: kleene-closure

    Perhaps that uv is already in L* because every concatenation of L is already in L* ?
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  4. #4
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    Re: kleene-closure

    Yes, if u and v are concatenations of words in L, then so is uv, and therefore it is in L*.
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