証明問題

$\displaystyle \sim (xy \lor \sim x z)$	$\displaystyle =$	$\displaystyle (\sim x \lor \sim y)(x \lor \sim z)$
	$\displaystyle =$	$\displaystyle \sim x x \lor \sim x \sim z \lor \sim y x \lor \sim y \sim z$
	$\displaystyle =$	$\displaystyle \sim x \sim z \lor \sim y x \lor \sim y \sim z$
	$\displaystyle =$	$\displaystyle \sim x \sim z (\sim y \lor y)$
		$\displaystyle \lor \sim y x (\sim z \lor z)$
		$\displaystyle \lor \sim y \sim z (\sim x \lor x)$
	$\displaystyle =$	$\displaystyle \sim x \sim z \sim y \lor \sim x \sim z y$
		$\displaystyle \lor \sim y x \sim z \lor \sim y x z$
		$\displaystyle \lor \sim y \sim z \sim x \lor \sim y \sim z x$
	$\displaystyle =$	$\displaystyle \sim x \sim y \sim z \lor \sim x y \sim z$
		$\displaystyle \lor x \sim y \sim z \lor x \sim y z$
		$\displaystyle \lor \sim x \sim y \sim z \lor x \sim y \sim z$
	$\displaystyle =$	$\displaystyle m_0 \lor m_2 \lor m_4 \lor m_5 \lor m_0 \lor m_4$
	$\displaystyle =$	$\displaystyle m_0 \lor m_2 \lor m_4 \lor m_5$

$\displaystyle \sim (xy \lor \sim x z)$	$\displaystyle =$	$\displaystyle (\sim x \lor \sim y)(x \lor \sim z)$
	$\displaystyle =$	$\displaystyle (\sim x \lor \sim y \lor \sim z z)(x \lor \sim z \lor \sim y y)$
	$\displaystyle =$	$\displaystyle (\sim x \lor \sim y \lor \sim z)(\sim x \lor \sim y \lor z)$
		$\displaystyle \land (x \lor \sim z \lor \sim y)(x \lor \sim z \lor y)$
	$\displaystyle =$	$\displaystyle (\sim x \lor \sim y \lor \sim z)(\sim x \lor \sim y \lor z)$
		$\displaystyle \land (x \lor \sim y \lor \sim z)(x \lor y \lor \sim z)$
	$\displaystyle =$	$\displaystyle M_0 M_1 M_4 M_6$

$\displaystyle \sim (xy \lor \sim x z)$	$\displaystyle =$	$\displaystyle M_0 M_1 M_4 M_6$
	$\displaystyle =$	$\displaystyle \sim \sim (M_0 M_1 M_4 M_6)$
	$\displaystyle =$	$\displaystyle \sim(\sim M_0 \lor \sim M_1 \lor \sim M_4 \lor \sim M_6)$
	$\displaystyle =$	$\displaystyle \sim(m_7 \lor m_6 \lor m_3 \lor m_1)$
	$\displaystyle =$	$\displaystyle m_0 \lor m_2 \lor m_4 \lor m_5$ $\displaystyle \mbox{$\because$\ 原関数に含まれないもの}$