<s>[INST] Convert a boolean expression to just NOR operatons How do we convert this expression to just $\text{NOR}$ operation?
$$A \cdot B + B \cdot C + C \cdot D$$
My attempt:
$A \cdot B + B \cdot C + C \cdot D = \overline{\overline{A \cdot B + B \cdot C + C \cdot D}} $
Using De Morgan's law
$ \to \overline{\overline{(A \cdot B)} \cdot \overline{(B \cdot C)} \cdot \overline{(C \cdot D)}} $
Using $X \text { NOR } Y = \overline{X} \cdot \overline{Y}$
$\to \overline{{(A \cdot B)} \text{ NOR } {(B \cdot C)} \text{ NOR } {(C \cdot D)}}$
Using $X \text { NOR } Y = \overline{X} \cdot \overline{Y}$
$\to \overline{{(\overline{A} \text{ NOR } \overline{B})} \text{ NOR } {(\overline{B} \text{ NOR } \overline{C})} \text{ NOR } {(\overline{C} \text{ NOR } \overline{D})}}$
Using $X \text { NOR } X = \overline{X} $
$\to \overline{((A \text{ NOR } A) \text{ NOR } (B \text{ NOR } B)) \text{ NOR } ((B \text{ NOR } B) \text{ NOR } (C \text{ NOR } C)) \text{ NOR } ((C \text{ NOR } C) \text{ NOR } (D \text{ NOR } D))}$
Using $X \text { NOR } X = \overline{X} $
$\to (((A \text{ NOR } A) \text{ NOR } (B \text{ NOR } B)) \text{ NOR } ((B \text{ NOR } B) \text{ NOR } (C \text{ NOR } C)) \text{ NOR } ((C \text{ NOR } C) \text{ NOR } (D \text{ NOR } D))) \text{ NOR } (((A \text{ NOR } A) \text{ NOR } (B \text{ NOR } B)) \text{ NOR } ((B \text{ NOR } B) \text{ NOR } (C \text{ NOR } C)) \text{ NOR } ((C \text{ NOR } C) \text{ NOR } (D \text{ NOR } D)))$
I am wondering if is there a better solution that return something simpler.
 [/INST] WolframAlpha returns
(a NOR c) NOR (b NOR c) NOR (b NOR d)

You can derive it by using the steps here.
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