Boolean algeabra prblem using T8/T8' distributivity theorems

[jennbjork]

Homework Statement

Convert each of the equations to an equivalent 2 - level implementation using t8/t8' distributivity theorems to get SOP/POS form. Simplify
F= Z((W'+Y+X')(W'+XY'Z))
G= ABC+D'(B+AC)(A'+C)'

Homework Equations

T8: x*(y+z)= x*y+x*z
T8': x+(y*z)=(x+y)*(x+z)

The Attempt at a Solution

I used T8' to take out w':
F=Z(W'+(X'+Y)(XY'Z))
=Z(W'+X'XY'Z+XYY'Z)
=Z(W'+0+0)
=Z*W


G= ABC+D'(B+AC)(A'+C)'

I distribute D' using T8:
ABC+(BD'+ACD')(A'D'+CD')'

but then do I continue as:
ABC+A'BD'D'+BC'D'D''+AA'CD'D'+ACC'D'D''
=ABC+A'BD'+BC'(0)+(0)CD'+A(0)(0)
=ABC+A'BD'+0+0 (x*x'=0; x*x=1; x*0=0)
=ABC+A'BD'

 

[member38644]

Or do I need to use T8' again to distribute the remaining terms? If so, how would I do that?

Your solution for F is correct, but for G, you can continue using T8' to distribute the remaining terms:

G= ABC+D'(B+AC)(A'+C')
= ABC+D'(B+A')(B+C')(A'+C') (using T8')
= ABC+D'(B+B'A'+BC'+A'C') (using T8')
= ABC+D'(B'+A'C') (using T8')
= ABC+D'B'+D'A'C' (using T8)

So the simplified SOP form for G is ABC+D'B'+D'A'C'.