# Convert the given Boolean function into its _NOR_ Normal Form.

Turn the Boolean function $$g(x)=x_0x_1 + x_0 \overline{x_2}+\overline{x_0}x_3+x_2x_3$$ into its NOR Normal Form (NOR - NF is when only NOR and NOT are used in the function).

Here's my attempt:

$$g(x)=x_0x_1 + x_0\overline{x_2}+\overline{x_0}x_3+x_2x_3 \\ = \overline{\overline{x_0x_1 + x_0\overline{x_2}+\overline{x_0}x_3+x_2x_3}} \\ = \overline{(\overline{x_0}+\overline{x_1}) \cdot (\overline{x_0}+x_2) \cdot (x_0+\overline{x_3}) \cdot (\overline{x_2}+\overline{x_3})} \\ = \overline{\overline{x_0}+ \overline{x_1}} + \overline{\overline{x_0}+ x_2} + \overline{x_0+ \overline{x_3}} + \overline{\overline{x_2}+ \overline{x_3}} \\ = \overline{\overline{\overline{\overline{x_0}+ \overline{x_1}} + \overline{\overline{x_0}+ x_2} + \overline{x_0+ \overline{x_3}} + \overline{\overline{x_2}+ \overline{x_3}}}}$$

I'm pretty sure this answer is correct since I only use NOT and NOR operations and if all the NOT operations are simplified using DeMorgan and Negation you get back the original function.

I don't understand the need for your last line. Your definition of "NOR normal form" is sightly vague, but in my opinion there is no doubt that the final answer should be your fourth line \begin{align} \overline{\overline{x_0}+ \overline{x_1}} + \overline{\overline{x_0}+ x_2} + \overline{x_0+ \overline{x_3}} + \overline{\overline{x_2}+ \overline{x_3}} \end{align} and not your fifth line \begin{align} \overline{\overline{\overline{\overline{x_0}+ \overline{x_1}} + \overline{\overline{x_0}+ x_2} + \overline{x_0+ \overline{x_3}} + \overline{\overline{x_2}+ \overline{x_3}}}} \end{align} Otherwise (i.e. if you accepted your fifth line as the "NOR normal form"), why not \begin{align} \overline{\overline{\overline{\overline{\overline{\overline{x_0}+ \overline{x_1}} + \overline{\overline{x_0}+ x_2} + \overline{x_0+ \overline{x_3}} + \overline{\overline{x_2}+ \overline{x_3}}}}}} \end{align} or \begin{align} \overline{\overline{ \overline{\overline{ \overline{\overline{\overline{\overline{x_0}+ \overline{x_1}} + \overline{\overline{x_0}+ x_2} + \overline{x_0+ \overline{x_3}} + \overline{\overline{x_2}+ \overline{x_3}}}}}}}} \end{align} ?
After the edit on the definition of NOR normal form: If a normal form is when only NOR and NOT can be used, then the right answer is your fifth line. Indeed, since $$\mathsf{nor}(a_1, \dots, a_n) = \overline{a_1 + \dots + a_n}$$, then \begin{align} &\overline{\overline{\overline{\overline{x_0}+ \overline{x_1}} + \overline{\overline{x_0}+ x_2} + \overline{x_0+ \overline{x_3}} + \overline{\overline{x_2}+ \overline{x_3}}}} \\[5pt] = \ & \overline{\overline{\mathsf{nor}(\overline{x_0}, \overline{x_1}) + \mathsf{nor}(\overline{x_0}, x_2) + \mathsf{nor}(x_0, \overline{x_3}) + \mathsf{nor}(\overline{x_2}, \overline{x_3})}} \\[5pt] = \ & \overline{\mathsf{nor} \big(\mathsf{nor}(\overline{x_0}, \overline{x_1}), \mathsf{nor}(\overline{x_0}, x_2), \mathsf{nor}(x_0, \overline{x_3}), \mathsf{nor}(\overline{x_2},\overline{x_3}) \big)} \end{align}
• Thanks for pointing it out. I realised I made a mistake in defining what the $NOR$ -NF is. I can only use $NOT$ and $NOR$ but not $OR$ operations. – Ski Mask Jun 4 at 8:48