Let $X$ be a D11: Set such that

(i) | $\mathcal{P}(X)$ is the D80: Power set of $X$ |

(ii) | $\{ 0, 1 \}^X$ is the D4213: Set of Boolean functions from $X$ to $\{ 0, 1 \}$ |

Then
\begin{equation}
\mathcal{P}(X)
\cong \{ 0, 1 \}^X
\end{equation}