x∈V
P(x)={y|y⊂x}
2^x={f:x→2}⊂x×2
g:2^x→P(x):g(f)={z∈x|(z,1)∈f}⊂x
h:P(x)→2^x:h(y)={(z,1)|z∈y}∪{(z,0)|z∈x-y}⊂2^x
B:boolean
B^x={f:x→B}⊂x×B
B^x→PB(x)
∪ p.o. ∪
2^x→P(x)
f∈B^x∩P(x)
f:x→B
f⊂x×B
f⊂x
f⊂x∩x×B=φ??