Is $f(X)f^d(X) = 0$ for a Boolean function $f$?

Question

I'm currently trying to understand a step in the proof for in the Crama and Hammer book on Boolean Functions. The proof is Proposition 4.12, which claims that the self-dualization of Boolean $f$ is bijective.

The specific statement that I am confused about is a step showing that:

$$f^d(X)f(X) \lor f(X) \bar{x}_{n+1} \lor f^d(X) {x}_{n+1} = f(X) \bar{x}_{n+1} \lor f^d(X) {x}_{n+1}.$$

Here $f(X)$ is a d-dimensional Boolean function, $X = (x_1,x_2,\ldots, x_n) \in \mathcal{B}^n$, and $f^d(X) := \overline{f(\overline{X})}$ is the dual of $f$.

This equality above seems to require that $f^d(X)f(X) = 0.$ I am wondering why this is? I think that I am missing something simple, or maybe I've found a typo in the proof. If someone has a different proof for the broader claim that self-dualization is bijective, that would be fine.

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I've copied the full proposition below in case it helps:

Here $f^{SD}(X)$ is the self-dual extension of $f$. This is a function over the (d+1)-dimensional Boolean hypercube such that $$f^{SD}(x_1,x_2,\ldots, x_n, x_{n+1}) = f(x_1,x_2,\ldots, x_n)\bar{x}_{n+1} \lor f^d(x_1,x_2,\ldots, x_n)\bar{x}_{n+1}.$$

Answer 1

$$f^d(X)f(X) \lor f(X) \bar{x}_{n+1} \lor f^d(X) {x}_{n+1} = f(X) \bar{x}_{n+1} \lor f^d(X) {x}_{n+1}$$ The equality above comes from the general equality below. It has nothing to do with the context.

$\begin{aligned} &\quad\ yz\lor y\overline{x_{n+1}} \lor z{x}_{n+1}\\ &=yz(\overline{x_{n+1}} \lor x_{n+1})\lor y\bar{x}_{n+1} \lor z{x}_{n+1}\\ &=(yz\overline{x_{n+1}} \lor yzx_{n+1})\lor y\bar{x}_{n+1} \lor z{x}_{n+1}\\ &=(yz\overline{x_{n+1}}\lor y\bar{x}_{n+1})\lor (yzx_{n+1}\lor z{x}_{n+1})\\ &=y\bar{x}_{n+1}\lor z{x}_{n+1}\\ \end{aligned}$