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

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

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

Any help?