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I did a logic network with only NAND gates to function below:

enter image description here

and my logic newtork

enter image description here

Is it correct?

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    \$\begingroup\$ Try marking each in-between output with the logical output to see if you have made an error i.e. break the problem down to smaller units. I can see an error but unless you start to analyse in sections you won't see that error. \$\endgroup\$
    – Andy aka
    Apr 17, 2020 at 8:53
  • \$\begingroup\$ Thank you for your answer. I think my problem is with x2 output and I'm not sure about this one inverter, I think I have to delete it and change also for nand. It will be ok? \$\endgroup\$
    – Lauro Mike
    Apr 17, 2020 at 9:04
  • \$\begingroup\$ Did you not understand my 1st comment and advice? \$\endgroup\$
    – Andy aka
    Apr 17, 2020 at 10:51
  • \$\begingroup\$ You need to show a bit more effort here. Take the Boolean equation and simplify it. From there it will be easier to spot any mistakes by doing the steps Andy suggests. \$\endgroup\$
    – MCG
    Apr 17, 2020 at 11:26
  • \$\begingroup\$ Unfortunately this function has been simplified. I can use only De Morgan laws to solve this. I've split this function for small steps and I did new circuit, I've uploaded it in Update. \$\endgroup\$
    – Lauro Mike
    Apr 17, 2020 at 16:04

2 Answers 2

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Your circuit simplifies to the equation Y = ~(~X1*X2)

enter image description here

While your equation seems to simplify to Y = ~(X1*X2*X3)

How I went about simplifying was to move the inversion to the Y side for the whole equation, to simplify the right hand side,

This leaves only getting a "1" when X1 = 0, X2 = 0 and the mess on the right, to get a 1 out of this immediately the second X2 can only be 0 and the second X1 can only be a 0 to have an output, meaning the only valid state for X3 is 0 to multiply to a change on the output.

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  • \$\begingroup\$ Thank you for your help. Unfortunately I can't simplify it more because I can use only De Morgan laws to solve this. \$\endgroup\$
    – Lauro Mike
    Apr 17, 2020 at 16:07
  • \$\begingroup\$ I think if we simplified the equation, it will NOT be Y = ~(X1*X2*X3), will it? \$\endgroup\$
    – NAND
    Apr 17, 2020 at 21:57
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First we should divide our problems to smaller parts.

We should begin from a symbol which has a lot of lines over it, in this case it's \$\overline{X_1}\$, but in the schematic there is no \$\overline{X_1}\$ found so it's first error.

Then we draw(track) \$\overline{\overline{X_1}X_3}\$, then we draw(track) \$\overline{X_2}\$ and so on to reach our function \$Y\$

Using these steps will make you analyze the problem and discover errors faster!.

I think schematic should be like that:

enter image description here

Here is the schematic after editing it.

enter image description here

But I think you'd better simplify the boolean equation algebraically then try to draw our schematic. If we simplify it we will get $$ Y=X_1+\overline{X_2}\,\overline{X_3} $$

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