boolean-logic - How to convert a 3 input AND gate into a NOR gate?

Question

I know that I can say convert a 2-input AND gate into a NOR gate by simply inverting the two inputs because of DeMorgan's Theorem.

But how would you do the equivalent on a 3-input AND gate?

Say...

     ____
A___|     \
B___|      )___ 
C___|____ /

I'm trying to understand this because my homework asks me to take a circuit and convert it using NOR synthesis to only use nor gates, and I know how to do it with 2 input gates, but the gate with 3 inputs is throwing me for a spin.

Answer 1

如果我没有犯任何错误，它几乎是一样的，反转所有 3 个输入，你会得到一个 NOR

桌子：

倒置的 AND 与

1 1 1 = 1           
1 1 1 = 0           
1 0 1 = 0           
0 1 0 = 0           
0 1 1 = 0           
0 1 0 = 0           
0 0 1 = 0           
0 0 0 = 0           

带有原始输入的 NOR

0 0 0 = 1            
0 0 1 = 0            
0 1 0 = 0            
1 0 1 = 0            
1 0 0 = 0            
1 0 1 = 0            
1 1 0 = 0            
1 1 1 = 0            

Answer 2

2 输入 AND 的 DeMorgan 定理将产生：

AB
(AB)''
(A' + B')'

所以，是的，输入被反转并馈入或非门。

DeMorgan 的 3 输入 AND 定理同样会产生：

ABC
(ABC)''
(A' + B' + C')'

同样，输入反相并馈入（3输入）或非门：

     ___
A--O\    \
B--O )    )O--- 
C--O/___ /

@SailorChibi 有显示等价的真值表。