I was given an expression which I simplified using k-maps to obtain the following:

~WY + W~YZ + ~W~X~Z

I followed along with some of the other workings people have done for converting functions, but I'm getting stuck. Here's an example:

~~(~WY + W~YZ + ~W~X~Z)


~( (W+~Y) + (~W + Y + ~Z) + (W + X + Z) )

My issue now is, where do I go from here? I have obtained some in direct NOR form, so for instance ~(W+~Y) is a nor expression. Issue is, I have to create a logic circuit for the final instance. And converting each of these to NOR, NOT = taking two inputs of the same literal into a single NOR gate, then taking W+~Y into a NOR gate as well.

Repeating this process for further terms leads to a similar issue where my final logic circuit is resulting in a lot of NOR gates that doesn't really simplify the circuit. If someone could confirm that my logic is correct I would appreciate it.

  • \$\begingroup\$ Hmm. Your last expression results in FALSE, though I think you are missing a closed paren. So I may be looking at it wrong. Using the double-NOT to start is the right beginning process, though. \$\endgroup\$ Commented Oct 4, 2023 at 14:50
  • \$\begingroup\$ I get ~( ~( ~W + ~(Y + ~Z) ) + ~( Y + ~( ~(~X + Z) + ~(W + ~Z) ) ) ). I can show the process, if needed. That's not the only approach, though. I had a choice at one stage where there were two directions to choose readily from. I picked one. The other would have created a different NOR result (same resulting logic, though.) \$\endgroup\$ Commented Oct 4, 2023 at 14:56
  • \$\begingroup\$ @periblepsis you're right I did miss out on the closed parenthesis, corrected that. If you could illustrate the main difference between the approach you took in the working versus mine I would appreciate it. I'm mostly still a bit iffy about the fact that, this simplification is a part of an earlier question, and somehow the NOR gates are supposed to be making the circuit more efficient. But if anything, this working only appears to be making it a more redundant circuit. \$\endgroup\$
    – agni_ka1
    Commented Oct 4, 2023 at 16:19
  • \$\begingroup\$ I don't have your context. All i can do is show a method that produces only 2-in NOR gates. \$\endgroup\$ Commented Oct 4, 2023 at 16:25

1 Answer 1


Well, here's my process:

$$\begin{align*} F&=\overline{W}Y + W\overline{Y}Z + \overline{W}\overline{X}\overline{Z} \\\\ &=\overline{\overline{\overline{W}Y + W\overline{Y}Z + \overline{W}\overline{X}\overline{Z}}} \\\\ &=\overline{W Y + W \overline{Z} + \overline{W}\overline{Y}Z+X\overline{Y}\overline{Z}} \\\\ &=\overline{\overline{\overline{W Y + W \overline{Z}}} + \overline{\overline{\overline{W}\overline{Y}Z+X\overline{Y}\overline{Z}}}} \\\\ &=\overline{\overline{\overline{W}+\overline{Y}Z} + \overline{Y+W Z + \overline{X}\overline{Z}}} \\\\ &=\overline{\overline{\overline{W}+\overline{Y+\overline{Z}}} + \overline{Y+\overline{\overline{W Z + \overline{X}\overline{Z}}}}} \\\\ &=\overline{\overline{\overline{W}+\overline{Y+\overline{Z}}} + \overline{Y+\overline{X \overline{Z}+\overline{W}Z}}} \\\\ &=\overline{\overline{\overline{W}+\overline{Y+\overline{Z}}} + \overline{Y+\overline{\overline{\overline{X}+ Z}+\overline{W+\overline{Z}}}}} \end{align*}$$

I've no promise that it's optimized in any way. And it's only for 2-in NORs. But it does follow a process that should work for you.

You seem already familiar with the basic idea. But just getting stuck here and there. Hopefully, the above clears up that log-jam.


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