Boolean Algebra Simplification

Question

How do I simplify the following equation? $\newcommand{\pn}{\phantom{\neg}}$ $$\begin{align*} \neg A\pn B \neg C \neg D\\ + \pn A\neg B\neg C\neg D\\ + \neg A\neg B\neg C\pn D\\ + \pn A\pn B\neg C\pn D\\ + \neg A\pn B\pn C\pn D\\ + \pn A\neg B\pn C\pn D\\ + \neg A\neg B\pn C\neg D\\ + \pn A\pn B\pn C\neg D \end{align*}$$

Answer 1

If the brackets are meant to be around all and-blocks, the statement is equivalent to $D$ being the parity bit for information $ABC$, i.e. $$D = \neg(A\oplus B \oplus C)\\ \Leftrightarrow A\oplus B \oplus C\oplus D$$ Where $\oplus$ is xor

Answer 2

Any propositional expression could be expressed by $1$ (true), $+$ (exclusive or) and $\cdot$ (and). Expressing an expression this way makes it straightforward to simplify by ordinary algebraic laws and calculations, just remembering that $S+S=0$ and $S\cdot S=S$ for all $S$.

1. $\neg S$ is replaced with $1+S$
2. $S\Rightarrow T$ is replaced with $1+S+S\cdot T$
3. $S\Leftrightarrow T$ is replaced with $1+S\cdot T$
4. $S \vee T$ is replaced with $S+T+S\cdot T$

In this case, this method would be clumsy, but it always work.