4-bit's 2's Complement using minimum number of Universal gates only?

Question

I'm trying to design a $4-$ bit's $2$' complement using minimum number of universal gates ONLY (NAND,NOR, NEGATIVE OR, NEGATIVE AND) , hence, minimum number of ICs.I got $4$ simplified output expressions ($w,x,y,z$) from the Truth table and k - map, and I constructed an individual circuits for each output, however, I'm ending up with 5 ICs(3NOR,2NAND), and I want to know if it's possible to minimiz the number of ICs to be less than 5 ICs. The output expressions are :

$W=A'(B+C+D)+AB'C'D'$
$X=B'(C+D)+BC'D'$
$Y=CD'+C'D$
$Z=D$

Thank you.

Answer 1

Your expressions are correct.

The following circuit can be implemented in four and a half ICs of the 74xx series:

8 NAND2 require 2   IC 7400 (quad NAND2)
6 NOR2  require 1,5 IC 7402 (quad NOR2)
6 INV   require 1   IC 7404 (hex INV)

The six inverters can be implemented by six NOR2 gates. This would result in 2 ICs 7400 and 3 ICs 7402 as described in the question.