化簡 $$$\overline{\overline{A \cdot B} + \left(\overline{D} \cdot A\right)}$$$

將德摩根定律 $$$\overline{x + y} = \overline{x} \cdot \overline{y}$$$ 應用於 $$$x = \overline{A \cdot B}$$$ 與 $$$y = \overline{D} \cdot A$$$：

$${\color{red}\left(\overline{\overline{A \cdot B} + \left(\overline{D} \cdot A\right)}\right)} = {\color{red}\left(\overline{\overline{A \cdot B}} \cdot \overline{\overline{D} \cdot A}\right)}$$

以 $$$x = A \cdot B$$$ 套用雙重否定（對合）律 $$$\overline{\overline{x}} = x$$$：

$${\color{red}\left(\overline{\overline{A \cdot B}}\right)} \cdot \overline{\overline{D} \cdot A} = {\color{red}\left(A \cdot B\right)} \cdot \overline{\overline{D} \cdot A}$$

將德摩根定律 $$$\overline{x \cdot y} = \overline{x} + \overline{y}$$$ 應用於 $$$x = \overline{D}$$$ 與 $$$y = A$$$：

$$A \cdot B \cdot {\color{red}\left(\overline{\overline{D} \cdot A}\right)} = A \cdot B \cdot {\color{red}\left(\overline{\overline{D}} + \overline{A}\right)}$$

以 $$$x = D$$$ 套用雙重否定（對合）律 $$$\overline{\overline{x}} = x$$$：

$$A \cdot B \cdot \left({\color{red}\left(\overline{\overline{D}}\right)} + \overline{A}\right) = A \cdot B \cdot \left({\color{red}\left(D\right)} + \overline{A}\right)$$

應用交換律：

$${\color{red}\left(A \cdot B \cdot \left(D + \overline{A}\right)\right)} = {\color{red}\left(A \cdot \left(D + \overline{A}\right) \cdot B\right)}$$

應用交換律：

$$A \cdot {\color{red}\left(D + \overline{A}\right)} \cdot B = A \cdot {\color{red}\left(\overline{A} + D\right)} \cdot B$$

將冗餘律$$$x \cdot \left(\overline{x} + y\right) = x \cdot y$$$應用於$$$x = A$$$與$$$y = D$$$：

$${\color{red}\left(A \cdot \left(\overline{A} + D\right)\right)} \cdot B = {\color{red}\left(A \cdot D\right)} \cdot B$$