1. ## boolean algebra

simplify the expression using boolean algebra: w'xy+wxz+wy'z+w'z' ?

2. ## Re: boolean algebra

If you need the final expression to be a CNF, then I don't think you can simplify much. The minimal number of conjuncts that I can see is three, versus four in the original formula. Also, it is difficult to do the simplification without a Karnaugh map.

Below, the first two steps split some conjuncts into two, and the rest of the steps recombine them in new ways. I write comments in curly braces.

xyw' + xzw + y'zw + z'w' = {xyw' = xy(z + z')w' = xyzw' + xyz'w'}
xyzw' + xyz'w' + xzw + y'zw + z'w' = {xzw = x(y + y')zw = xyzw + xy'zw}
xyzw' + xyz'w' + xyzw + xy'zw + y'zw + z'w' = {xyz'w' + z'w' = (xy + 1)z'w' = z'w'}
xyzw' + xyzw + xy'zw + y'zw + z'w' = {xy'zw + y'zw = (x + 1)y'zw = y'zw}
xyzw' + xyzw + y'zw + z'w' = {xyzw' + xyzw = xyz(w + w') = xyz}
xyz + y'zw + z'w'

Edit: It is recommended to post questions about Boolean algebra in the Discrete Math subforum.

3. ## Re: boolean algebra

The NPMACS.COM has opened a free tools for solving the simplification of logical function. It is totally free.

You can get the answer immediately:
\$f=
w*!y*z
+
x*y*z
+
!w*!z