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Math Help - Indentification between X and the truth valuation of its boolean function

  1. #1
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    Indentification between X and the truth valuation of its boolean function

    Hey guys, I'm kind of stuck here. Any hint will be of great help

    Let X be a finite set and FB(X) the boolean algebra of boolean functions of X. Im trying to show that there is an identification between X and the truth valuation of the boolean functions if X is finite. In this case a truth valuation is a morphism between FB(X) (all functions from X to  \left \{ 0,1 \right \} ) and  \left \{ 0,1 \right \} . To proof this im going to use the function v_{x}(f)=f(x)   \forall f \in FB(X) and proof its bijective.

    I proved that v_{x} is one to one, but im having a hard time proving onto. I took a function g:FB(X) \rightarrow  \left \{ 0,1 \right \} and then the set Y = \left \{ v_{1},...,v_{s} \in FB(X) | g(v_{i})= 0 \right \}. I then wanted to used the fact that the sets P_{v_{i}} = \left \{ x \in X | v_{i}(x)=0 \right \} are generated by a single element (a_{i})to then construct a =  \prod_{i} a_{i} so that v_{a} = g. But realized that since X is not an algebra I cant take the product and also that since the elements of FB(X) are not morphism, I cant guarantee that P_{v_{i}} is not empty.

    Any help?
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  2. #2
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    Re: Indentification between X and the truth valuation of its boolean function

    Quote Originally Posted by gordo151091 View Post
    Let X be a finite set and FB(X) the boolean algebra of boolean functions of X.
    A Boolean function takes truth values (0 or 1) as argument, not elements of an arbitrary set X.

    Quote Originally Posted by gordo151091 View Post
    Im trying to show that there is an identification between X and the truth valuation of the boolean functions if X is finite.
    A single truth valuation? (You are saying "the" and "valuation" in singular.) What do you mean by identification between a set X and a function (since a truth valuation is a function)?

    Quote Originally Posted by gordo151091 View Post
    In this case a truth valuation is a morphism between FB(X) (all functions from X to  \left \{ 0,1 \right \} ) and  \left \{ 0,1 \right \} . To proof this im going to use the function v_{x}(f)=f(x)   \forall f \in FB(X) and proof its bijective.

    I proved that v_{x} is one to one
    Do you mean $v_x$ for a concrete $x$ or $v$ as a function of $x$? The function $v_x$ is from $FB(X)$ to {0, 1} and is certainly not one-to-one.

    Quote Originally Posted by gordo151091 View Post
    but im having a hard time proving onto.
    One frequently denotes {0, 1} by 2 and the set of functions from $A$ to $B$ by $B^A$, since the number of elements in $B^A$ is $|B|^{|A|}$, where $|Z|$ denotes the number of elements in some $Z$. Then $FB(X)=2^X$. Are you trying to prove that $v$ is a bijection from $X$ to $2^{2^X}$? This cannot happen because $2^{2^{|X|}}$ is much larger than $|X|$.
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