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- \documentclass{article}
- \usepackage[utf8]{inputenc}
- \usepackage[portuguese]{babel}%
- \usepackage[shortlabels]{enumitem}
- \usepackage{amsfonts}
- \usepackage{amsmath}
- \usepackage{amssymb}
- \usepackage{setspace}
- \doublespacing
- \newtheorem{lemma}{Lema}
- \newtheorem{thm}{Teorema}
- \newtheorem{corollary}{Corolário}
- \newtheorem{definition}{Definição}
- \usepackage[margin=2cm]{geometry}
- \usepackage[divipsnames]{xcolor}
- \newenvironment{proof}[1][Prova]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}}
- \title{Lista 1 - MAC0329}
- \author{Gustavo Korzune Gurgel\\No USP: 4778350}
- \date{Abril 2020}
- \begin{document}
- \maketitle
- Irei demonstrar que a sêxtupla $\langle B^n,\,+,\,\cdot,\,-,\,\mathbf{0},\,\mathbf{1} \rangle$ é uma álgebra booleana por meio dos axiomas da álgebra booleana.
- \begin{proof}
- A partir das próprias operações e identidades de $\langle B^n,\,+,\,\cdot,\,-,\,\mathbf{0},\,\mathbf{1} \rangle$ definidas nas notas de aula, das operações e identidades $\langle B,\,\textcolor{magenta}{+},\,\textcolor{magenta}{\cdot},\,\textcolor{magenta}{-},\,\textcolor{magenta}{0},\,\textcolor{magenta}{1} \rangle$ e do fato que $\langle B,\,\textcolor{magenta}{+},\,\textcolor{magenta}{\cdot},\,\textcolor{magenta}{-},\,\textcolor{magenta}{0},\,\textcolor{magenta}{1} \rangle$ é uma álgebra booleana faremos todos os argumentos dessa prova.\\
- Segue-se agora a prova de que os quatro axiomas da álgebra booleana são satisfeitos por $\langle B^n,\,+,\,\cdot,\,-,\,\mathbf{0},\,\mathbf{1} \rangle$.
- \begin{enumerate}[1.]
- \item \underline{As operações $+$ e $\cdot$ são comutativas:} \footnote{Repare que na Eq. \eqref{comut+} e Eq. \eqref{comut.} usamos, respectivamente, o fato que a operação $\textcolor{magenta}{+}$ é comutativa e que a operação $\textcolor{magenta}{\cdot}$ é comutativa.}
- \begin{equation}
- \label{comut+}
- \begin{split}
- (x_1, \;x_2, \;\dots, \;x_{n}) + (y_1,\;y_2,\;\dots,\;y_{n}) = (x_1 \; \textcolor{magenta}{+} \; y_1, \; x_2 \; \textcolor{magenta}{+} \; y_2, \; \dots,\; x_{n} \; \textcolor{magenta}{+} \; y_{n})\\
- = (y_1 \; \textcolor{magenta}{+} \; x_1, \; y_2 \; \textcolor{magenta}{+} \; x_2, \; \dots, \; y_{n} \; \textcolor{magenta}{+} \; x_{n}) \\
- = (y_1,\;y_2,\;\dots,\;y_{n}) + (x_1, \;x_2, \;\dots, \;x_{n})
- \end{split}
- \end{equation}
- Portanto, $$ (x_1, \;x_2, \;\dots, \;x_{n}) + (y_1,\;y_2,\;\dots,\;y_{n}) = (y_1,\;y_2,\;\dots,\;y_{n}) + (x_1, \;x_2, \;\dots, \;x_{n}) $$
- Que, por definição, significa que $+$ é comutativa. \checkmark
- \begin{equation}
- \label{comut.}
- \begin{split}
- (x_1, \;x_2, \;\dots, \;x_{n}) \cdot (y_1,\;y_2,\;\dots,\;y_{n}) = (x_1 \; \textcolor{magenta}{\cdot} \; y_1, \; x_2 \; \textcolor{magenta}{\cdot} \; y_2, \; \dots, \; x_{n} \; \textcolor{magenta}{\cdot} \; y_{n}) \\
- = (y_1 \; \textcolor{magenta}{\cdot} \; x_1, \; y_2 \; \textcolor{magenta}{\cdot} \; x_2, \; \dots , \; y_{n} \; \textcolor{magenta}{\cdot} \; x_{n})\\
- = (y_1,\;y_2,\;\dots,\;y_{n}) \cdot (x_1, \;x_2, \;\dots, \;x_{n})
- \end{split}
- \end{equation}
- Portanto, \newpage
- $$ (x_1, \;x_2, \;\dots, \;x_{n}) \cdot (y_1,\;y_2,\;\dots,\;y_{n}) = (y_1,\;y_2,\;\dots,\;y_{n}) \cdot (x_1, \;x_2, \;\dots, \;x_{n}) $$
- Que, por definição, significa que $\cdot$ é comutativa. \checkmark
- Sendo assim, $+$ e $\cdot$ são comutativas. \qquad $\square$
- \item \underline{$\cdot$ é distributiva em $+$, e, $+$ é distributiva em $\cdot$ :}
- \begin{equation*}
- \begin{split}
- (x_1, \;x_2, \;\dots, \;x_{n}) \cdot \big((y_1,\;y_2,\;\dots,\;y_{n}) + (z_1, \;z_2, \;\dots, \;z_{n})\big) = \\
- (x_1, \;x_2, \;\dots, \;x_{n}) \cdot (y_1 \; \textcolor{magenta}{+} \; z_1, \; y_2 \; \textcolor{magenta}{+} \; z_2, \; \dots, \; y_{n} \; \textcolor{magenta}{+} \; z_{n}) = \\
- \big(x_1 \textcolor{magenta}{\cdot} (y_1 \; \textcolor{magenta}{+} \; z_1), \; x_2 \textcolor{magenta}{\cdot} (y_2 \; \textcolor{magenta}{+} \; z_2), \; \dots, \; x_{n} \textcolor{magenta}{\cdot} (y_{n} \; \textcolor{magenta}{+} \; z_{n})\big)
- \end{split}
- \end{equation*}
- Sabemos das notas de aula que $\textcolor{magenta}{\cdot}$ é distributiva em $\textcolor{magenta}{+}$, e sabemos também que \\ $\forall i \in \{ 1, 2, ..., n \} (x_i \in B \, \wedge \, y_i \in B \, \wedge \, z_i \in B)$. Portanto,
- \begin{equation*}
- \begin{split}
- \big(x_1 \; \textcolor{magenta}{\cdot} \; (y_1 \; \textcolor{magenta}{+} \; z_1), \; x_2 \; \textcolor{magenta}{\cdot} \; (y_2 \; \textcolor{magenta}{+} \; z_2), \; \dots, \; x_{n} \; \textcolor{magenta}{\cdot} \; (y_{n} \; \textcolor{magenta}{+} \; z_{n})\big) = \\
- \big((x_1 \; \textcolor{magenta}{\cdot} \; y_1) \; \textcolor{magenta}{+} \; (x_1 \; \textcolor{magenta}{\cdot} \; z_1), \; (x_2 \; \textcolor{magenta}{\cdot} \; y_2) \; \textcolor{magenta}{+} \; (x_2 \; \textcolor{magenta}{\cdot} \; z_2), \; \dots, \; (x_{n} \; \textcolor{magenta}{\cdot} \; y_{n}) \; \textcolor{magenta}{+} \; (x_{n} \; \textcolor{magenta}{\cdot} \; z_{n})\big) = \\
- (x_1 \; \textcolor{magenta}{\cdot} \; y_1, \; x_2 \; \textcolor{magenta}{\cdot} \; y_2, \; \dots, \; x_{n} \; \textcolor{magenta}{\cdot} \; y_{n}) + (x_1 \; \textcolor{magenta}{\cdot} \; z_1, \; x_2 \; \textcolor{magenta}{\cdot} \; z_2, \; \dots, \; x_{n} \; \textcolor{magenta}{\cdot} \; z_{n}) =\\
- \big((x_1, \;x_2, \;\dots, \;x_{n}) \cdot (y_1,\;y_2,\;\dots,\;y_{n})\big) + \big((x_1, \;x_2, \;\dots, \;x_{n}) \cdot (z_1,\;z_2,\;\dots,\;z_{n})\big)
- \end{split}
- \end{equation*}
- Portanto,
- \begin{equation*}
- \begin{split}
- (x_1, \;x_2, \;\dots, \;x_{n}) \cdot \big((y_1,\;y_2,\;\dots,\;y_{n}) + (z_1, \;z_2, \;\dots, \;z_{n})\big) = \\
- \big((x_1, \;x_2, \;\dots, \;x_{n}) \cdot (y_1,\;y_2,\;\dots,\;y_{n})\big) + \big((x_1, \;x_2, \;\dots, \;x_{n}) \cdot (z_1,\;z_2,\;\dots,\;z_{n})\big)
- \end{split}
- \end{equation*}
- Que, por definição, significa que $\cdot$ é distributiva em $+$ \checkmark
- \begin{equation*}
- \begin{split}
- (x_1, \;x_2, \;\dots, \;x_{n}) + \big((y_1,\;y_2,\;\dots,\;y_{n}) \cdot (z_1, \;z_2, \;\dots, \;z_{n})\big) = \\
- (x_1, \;x_2, \;\dots, \;x_{n}) + (y_1 \; \textcolor{magenta}{\cdot} \; z_1, \; y_2 \; \textcolor{magenta}{\cdot} \; z_2, \; \dots, \; y_{n} \; \textcolor{magenta}{\cdot} \; z_{n}) = \\
- \big(x_1 \; \textcolor{magenta}{+} \; (y_1 \; \textcolor{magenta}{\cdot} \; z_1), \; x_2 \; \textcolor{magenta}{+} \; (y_2 \; \textcolor{magenta}{\cdot} \; z_2), \; \dots, \; x_{n} \; \textcolor{magenta}{+} \; (y_{n} \; \textcolor{magenta}{\cdot} \; z_{n})\big)
- \end{split}
- \end{equation*}
- Sabemos das notas de aula que $\textcolor{magenta}{+}$ é distributiva em $\textcolor{magenta}{\cdot}$, e sabemos também que \\ $\forall i \in \{ 1, 2, ..., n \} (x_i \in B \, \wedge \, y_i \in B \, \wedge \, z_i \in B)$. Portanto,
- \begin{equation*}
- \begin{split}
- \big(x_1 \; \textcolor{magenta}{+} \; (y_1 \; \textcolor{magenta}{\cdot} \; z_1), \; x_2 \; \textcolor{magenta}{+} \; (y_2 \; \textcolor{magenta}{\cdot} \; z_2), \; \dots, \; x_{n} \; \textcolor{magenta}{+} \; (y_{n} \; \textcolor{magenta}{\cdot} \; z_{n})\big) = \\
- \big((x_1 \; \textcolor{magenta}{+} \; y_1) \; \textcolor{magenta}{\cdot} \; (x_1 \; \textcolor{magenta}{+} \; z_1), \; (x_2 \; \textcolor{magenta}{+} \; y_2) \; \textcolor{magenta}{\cdot} \; (x_2 \; \textcolor{magenta}{+} \; z_2), \; \dots, \; (x_{n} \; \textcolor{magenta}{+} \; y_{n}) \; \textcolor{magenta}{\cdot} \; (x_{n} \; \textcolor{magenta}{+} \; z_{n})\big) = \\
- (x_1 \; \textcolor{magenta}{+} \; y_1, \; x_2 \; \textcolor{magenta}{+} \; y_2, \; \dots, \; x_{n} \; \textcolor{magenta}{+} \; y_{n}) \cdot (x_1 \; \textcolor{magenta}{+} \; z_1, \; x_2 \; \textcolor{magenta}{+} \; z_2, \; \dots, \; x_{n} \; \textcolor{magenta}{+} \; z_{n}) =\\
- \big((x_1, \;x_2, \;\dots, \;x_{n}) + (y_1,\;y_2,\;\dots,\;y_{n})\big) \cdot \big((x_1, \;x_2, \;\dots, \;x_{n}) + (z_1,\;z_2,\;\dots,\;z_{n})\big)
- \end{split}
- \end{equation*}
- Portanto,
- \begin{equation*}
- \begin{split}
- (x_1, \;x_2, \;\dots, \;x_{n}) + \big((y_1,\;y_2,\;\dots,\;y_{n}) \cdot (z_1, \;z_2, \;\dots, \;z_{n})\big) = \\
- \big((x_1, \;x_2, \;\dots, \;x_{n}) + (y_1,\;y_2,\;\dots,\;y_{n})\big) \cdot \big((x_1, \;x_2, \;\dots, \;x_{n}) + (z_1,\;z_2,\;\dots,\;z_{n})\big)
- \end{split}
- \end{equation*}
- Que, por definição, significa que $+$ é distributiva em $\cdot$ \checkmark
- \\
- Sendo assim, $\cdot$ é distributiva em $+$, e, $+$ é distributiva em $\cdot$ $\quad \square$
- \item \underline{\textbf{0} e \textbf{1} são identidade:}
- $$ (x_1, \;x_2, \;\dots, \;x_{n}) + \mathbf{0} = (x_1, \;x_2, \;\dots, \;x_{n}) + (\textcolor{magenta}{0},\; \textcolor{magenta}{0}, \; \dots, \; \textcolor{magenta}{0}) = (x_1 \; \textcolor{magenta}{+} \; \textcolor{magenta}{0}, \; x_2 \; \textcolor{magenta}{+} \; \textcolor{magenta}{0} ,\;\dots , \;x_{n} \; \textcolor{magenta}{+} \; \textcolor{magenta}{0}) $$
- $\forall i \in \{ 1,2, \dots, n \}\big( x_i \in B \big) \implies \forall i \in \{ 1,2, \dots, n \} \big(x_i \; \textcolor{magenta}{+} \; \textcolor{magenta}{0} = x_i\big) $. \\
- $\therefore (x_1 \; \textcolor{magenta}{+} \; \textcolor{magenta}{0}, \; x_2 \; \textcolor{magenta}{+} \; \textcolor{magenta}{0} ,\;\dots , \;x_{n} \; \textcolor{magenta}{+} \; \textcolor{magenta}{0}) = (x_1, \; x_2 ,\;\dots , \;x_{n})$ e, portanto, $(x_1, \; x_2 ,\;\dots , \;x_{n}) + \textbf{0} = (x_1, \; x_2 ,\;\dots , \;x_{n})$, em outras palavras, \textbf{0} é identidade. \checkmark
- \\
- $$ (x_1, \;x_2, \;\dots, \;x_{n}) \cdot \mathbf{1} = (x_1, \;x_2, \;\dots, \;x_{n}) \cdot (\textcolor{magenta}{1},\; \textcolor{magenta}{1}, \; \dots, \; \textcolor{magenta}{1}) = (x_1 \; \textcolor{magenta}{\cdot} \; \textcolor{magenta}{1}, \; x_2 \; \textcolor{magenta}{\cdot} \; \textcolor{magenta}{1} ,\;\dots , \;x_{n} \; \textcolor{magenta}{\cdot} \; \textcolor{magenta}{1})$$
- $\forall i \in \{ 1,2, \dots, n \}\big( x_i \in B \big) \implies \forall i \in \{ 1,2, \dots, n \} \big(x_i \; \textcolor{magenta}{\cdot} \; \textcolor{magenta}{1} = x_i\big) $. \\
- $\therefore (x_1 \; \textcolor{magenta}{\cdot} \; \textcolor{magenta}{1}, \; x_2 \; \textcolor{magenta}{\cdot} \; \textcolor{magenta}{1} ,\;\dots , \;x_{n} \; \textcolor{magenta}{\cdot} \; \textcolor{magenta}{1}) = (x_1, \; x_2 ,\;\dots , \;x_{n})$ e, portanto, $(x_1, \; x_2 ,\;\dots , \;x_{n}) \cdot \textbf{1} = (x_1, \; x_2 ,\;\dots , \;x_{n})$, em outras palavras, \textbf{1} é identidade. \checkmark
- Em suma, conclui-se que \textbf{0} e \textbf{1} são identidade. \quad $\square$
- \item \underline{$\overline{(x_1, \;x_2, \;\dots, \;x_{n})}$ é complemento de $(x_1, \;x_2, \;\dots, \;x_{n})$:}
- \begin{equation*}
- \begin{split}
- \overline{(x_1, \;x_2, \;\dots, \;x_{n})} + (x_1, \;x_2, \;\dots, \;x_{n}) = (\textcolor{magenta}{\overline{\textcolor{black}{x}}}_1, \; \textcolor{magenta}{\overline{\textcolor{black}{x}}}_2, \;\dots, \; \textcolor{magenta}{\overline{\textcolor{black}{x}}}_{n}) + (x_1, \;x_2, \;\dots, \;x_{n}) \\
- = (\textcolor{magenta}{\overline{\textcolor{black}{x}}}_1 \; \textcolor{magenta}{+} \; x_1, \; \textcolor{magenta}{\overline{\textcolor{black}{x}}}_2 \; \textcolor{magenta}{+} \; x_2 , \;\dots, \; \textcolor{magenta}{\overline{\textcolor{black}{x}}}_{n} \; \textcolor{magenta}{+} \; x_n )
- \end{split}
- \end{equation*}
- $\forall i \in \{ 1,2, \dots, n \}\big( \textcolor{magenta}{\overline{\textcolor{black}{x}}}_{i} \; \textcolor{magenta}{+} \; x_i = 1 \big) \Rightarrow (\textcolor{magenta}{\overline{\textcolor{black}{x}}}_1 \; \textcolor{magenta}{+} \; x_1, \; \textcolor{magenta}{\overline{\textcolor{black}{x}}}_2 \; \textcolor{magenta}{+} \; x_2 , \;\dots, \; \textcolor{magenta}{\overline{\textcolor{black}{x}}}_{n} \; \textcolor{magenta}{+} \; x_n ) = (\textcolor{magenta}{1},\; \textcolor{magenta}{1}, \; \dots, \; \textcolor{magenta}{1}) = \textbf{1}.$\\
- $\therefore \overline{(x_1, \;x_2, \;\dots, \;x_{n})} + (x_1, \;x_2, \;\dots, \;x_{n}) = \textbf{1}$ \checkmark
- \begin{equation*}
- \begin{split}
- \overline{(x_1, \;x_2, \;\dots, \;x_{n})} \cdot (x_1, \;x_2, \;\dots, \;x_{n}) = (\textcolor{magenta}{\overline{\textcolor{black}{x}}}_1, \; \textcolor{magenta}{\overline{\textcolor{black}{x}}}_2, \;\dots, \; \textcolor{magenta}{\overline{\textcolor{black}{x}}}_{n}) \cdot (x_1, \;x_2, \;\dots, \;x_{n}) \\
- = (\textcolor{magenta}{\overline{\textcolor{black}{x}}}_1 \; \textcolor{magenta}{\cdot} \; x_1, \; \textcolor{magenta}{\overline{\textcolor{black}{x}}}_2 \; \textcolor{magenta}{\cdot} \; x_2 , \;\dots, \; \textcolor{magenta}{\overline{\textcolor{black}{x}}}_{n} \; \textcolor{magenta}{\cdot} \; x_n )
- \end{split}
- \end{equation*}
- $\forall i \in \{ 1,2, \dots, n \}\big( \textcolor{magenta}{\overline{\textcolor{black}{x}}}_{i} \; \textcolor{magenta}{\cdot} \; x_i = 0 \big) \Rightarrow (\textcolor{magenta}{\overline{\textcolor{black}{x}}}_1 \; \textcolor{magenta}{\cdot} \; x_1, \; \textcolor{magenta}{\overline{\textcolor{black}{x}}}_2 \; \textcolor{magenta}{\cdot} \; x_2 , \;\dots, \; \textcolor{magenta}{\overline{\textcolor{black}{x}}}_{n} \; \textcolor{magenta}{\cdot} \; x_n ) = (\textcolor{magenta}{0},\; \textcolor{magenta}{0}, \; \dots, \; \textcolor{magenta}{0}) = \textbf{0}.$\\
- $\therefore \overline{(x_1, \;x_2, \;\dots, \;x_{n})} + (x_1, \;x_2, \;\dots, \;x_{n}) = \textbf{0}$ \checkmark
- Logo, $\overline{(x_1, \;x_2, \;\dots, \;x_{n})}$ é complemento de $(x_1, \;x_2, \;\dots, \;x_{n}). \qquad \square $
- \end{enumerate}{}
- Por fim, uma vez que os quatro axiomas da álgebra booleana mostraram-se satisfeitos para a sêxtupla $\langle B^n,\,+,\,\cdot,\,-,\,\mathbf{0},\,\mathbf{1} \rangle$ com as operações e elementos definidos tal qual o enunciado da lista, pode-se concluir que, de fato, $\langle B^n,\,+,\,\cdot,\,-,\,\mathbf{0},\,\mathbf{1} \rangle$ é uma álgebra booleana.
- \end{proof}
- \end{document}
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