CHAP. 15] BOOLEAN ALGEBRA 393BOOLEAN EXPRESSIONS15.10. Reduce the following Boolean products to either 0 or a fundamental product:
(a) xyx′z; (b) xyzy; (c) xyz′yx; (d) xyz′yx′z′Use the commutative lawx∗y=y∗x, the complement lawx∗x′=0, and the idempotent lawx∗x=x:(a) xyx′z=xx′yz= 0 yz= 0
(b) xyzy=xyyz=xyz
(c) xyz′yx=xxyyz′=xyz′
(d) xyz′yx′z′=xx′yyz′z′= 0 yz′= 015.11. ExpresseachBooleanexpressionE(x, y, z)asasum-of-productsandtheninitscompletesum-of-products
form: (a) E=x(xy′+x′y+y′z); (b) E=z(x′+y)+y′.
First use Algorithm 15.1 to expressEas a sum-of-products, and then use Algorithm 15.2 to expressEas a
complete sum-of-products.
(a) First we haveE=xxy′+xx′y+xy′z=xy′+xy′z. Then
E=xy′(z+z′)+xy′z=xy′z+xy′z′+xy′z=xy′z+xy′z′(b) First we have
E=z(x′+y)+y′=x′z+yz+y′
Then
E=x′z+yz+y′=x′z(y+y′)+yz(x+x′)+y′(x+x′)(z+z′)
=x′yz+x′y′z+xyz+x′yz+xy′z+xy′z′+x′y′z+x′y′z′
=xyz+xy′z+xy′z′+x′yz+x′y′z+x′y′z′15.12. ExpressE(x, y, z)=(x′+y)′+x′yin its complete sum-of-products form.
We haveE=(x′+y)′+x′y=xy′+x′y, which would be the complete sum-of-products form ofEifEwere
a Boolean expression inxandy. However, it is specified thatEis a Boolean expression in the three variablesx,y,
andz. Hence,E=xy′+x′y=xy′(z+z′)+x′y(z+z′)=xy′z+xy′z′+x′yz+x′yz′is the complete sum-of-products form ofE.15.13. ExpresseachBooleanexpressionE(x, y, z)asasum-of-productsandtheninitscompletesum-of-products
form: (a) E=y(x+yz)′; (b) E=x(xy+y′+x′y).
(a) E=y(x′(yz)′)=yx′(y′+z′)=yx′y′+x′yz′=x′yz′
which already is in its complete sum-of-products form.
(b) First we haveE=xxy+xy′+xx′y=xy+xy′. Then
E=xy(z+z′)+xy′(z+z′)=xyz+xyz′+xy′z+xy′z′15.14. Express each set expressionE(A, B, C)involving setsA,B,Cas a union of intersections:
(a) E=(A∪B)c∩(Cc∪B); (b) E=(B∩C)c∩(Ac∩C)cUse Boolean notation,′for complement,+for union, and∗(or juxtaposition) for intersection, and then express
Eas a sum of products (union of intersections).(a) E=(A+B)′(C′+B)=A′B′(C′+B)=A′B′C′+A′B′B=A′B′C′orE=Ac∩Bc∩Cc
(b) E=(BC)′(A′+C)′=(B′+C′)(AC′)=AB′C′+AC′orE=(A∩Bc∩Cc)∩(A∩Cc)