# Answer to Question #2474 in Discrete Mathematics for Rob

Question #2474

Show how you derive the sum-of-products expansion of the Boolean function f(x,y,z) = x(~y + ~yz + ~x) + (~xy + y~z)(x~yz + ~xz).

Note* - ~ denotes the complement of an element

ex- ~0 = 1 and ~1 = 0

Note* - ~ denotes the complement of an element

ex- ~0 = 1 and ~1 = 0

Expert's answer

Notice that

x ~x = 0, x x = x,& 1+x = 1

Then

f(x,y,z) = x(~y+~yz+~x)+(~xy+y~z)(x~yz+~xz) =

=& x~y+x~yz+x~x + ~xyx~yz + ~xy~xz + y~zx~yz +~ y~zxz =

= x~y+x~yz+ 0 + 0 + ~xyz + 0 + 0 =

= x~y + x~yz + ~xyz =

= x~y(1+z) + ~xyz =

= x~y + ~xyz.

x ~x = 0, x x = x,& 1+x = 1

Then

f(x,y,z) = x(~y+~yz+~x)+(~xy+y~z)(x~yz+~xz) =

=& x~y+x~yz+x~x + ~xyx~yz + ~xy~xz + y~zx~yz +~ y~zxz =

= x~y+x~yz+ 0 + 0 + ~xyz + 0 + 0 =

= x~y + x~yz + ~xyz =

= x~y(1+z) + ~xyz =

= x~y + ~xyz.

## Comments

## Leave a comment