化簡 $$$0 \oplus 1$$$

將 $$$x = 0$$$ 與 $$$y = 1$$$ 代入公式 $$$x \oplus y = \left(x \cdot \overline{y}\right) + \left(\overline{x} \cdot y\right)$$$：

$${\color{red}\left(0 \oplus 1\right)} = {\color{red}\left(\left(0 \cdot \overline{1}\right) + \left(\overline{0} \cdot 1\right)\right)}$$

套用否定律 $$$\overline{1} = 0$$$：

$$\left(0 \cdot {\color{red}\left(\overline{1}\right)}\right) + \left(\overline{0} \cdot 1\right) = \left(0 \cdot {\color{red}\left(0\right)}\right) + \left(\overline{0} \cdot 1\right)$$

套用否定律 $$$\overline{0} = 1$$$：

$$\left(0 \cdot 0\right) + \left({\color{red}\left(\overline{0}\right)} \cdot 1\right) = \left(0 \cdot 0\right) + \left({\color{red}\left(1\right)} \cdot 1\right)$$

對 $$$x = 0$$$ 應用支配（零、歸零）律 $$$x \cdot 0 = 0$$$：

$${\color{red}\left(0 \cdot 0\right)} + \left(1 \cdot 1\right) = {\color{red}\left(0\right)} + \left(1 \cdot 1\right)$$

應用交換律：

$${\color{red}\left(0 + \left(1 \cdot 1\right)\right)} = {\color{red}\left(\left(1 \cdot 1\right) + 0\right)}$$

將恆等律 $$$x + 0 = x$$$ 應用於 $$$x = 1 \cdot 1$$$：

$${\color{red}\left(\left(1 \cdot 1\right) + 0\right)} = {\color{red}\left(1 \cdot 1\right)}$$

將恆等律 $$$x \cdot 1 = x$$$ 應用於 $$$x = 1$$$：

$${\color{red}\left(1 \cdot 1\right)} = {\color{red}\left(1\right)}$$