1. Do problem 18 on page 112 at the end of Chapter 3.
a. Is the lattice complemented?
NO. Counter-example: There does not
exist a x such that ax=0 and a+x=1.
b. Is the lattice distributed?
NO. Counter-example: d(b+f)=d 1=d but
db+df = 1+1=1
c. Is the lattice a Boolean Algebra?
NO. It must be both complemented
and distributed to be a Boolean Algebra.
x y f(x,y) 0 0 1 0 1 0 1 0 0 1 1 1 Minterm canonical form: f(x,y) = x'y' + xy Maxterm canonical form: f(x,y) = (1 + x + y) (0+x + y')(0+x'+y)(1+x + y) = (x + y')(x'+y) since 1+z = 1 for every z in B.b. f(x,y)= ax + by
x y f(x,y) 0 0 0 0 1 b 1 0 a 1 1 a+b Minterm canonical form: f(x,y) = bx'y + axy' +(a+b)xy Maxterm canonical form: f(x,y) = (x+y)(b+x+y')(a+x'+y)(a+b+x'+y')c. f(x,y,z) = axy +a'(y+z') + xz
x y z f(x,y,z)
0 0 0 a'
0 0 1 0
0 1 0 a'
0 1 1 a'
1 0 0 a'
1 0 1 1
1 1 0 1
1 1 1 1
Minterm Canonical Form: f(x,y,z)= a'x'y'z' + a'x'yz' + a'x'yz + a'xy'z' +
xy'z + xyz' + xyz
Maxterm Canonical Form:
f(x,y,z) =
(a'+x+y+z)(0+x+y+z')(a'+x+y'+z)(a'+x+y'+z')(a'+x'+y+z)(1+x'+y+z')(1+x'+y'+z)(1+x'+y'+z')
= (a'+x+y+z)( x+y+z')(a'+x+y'+z)(a'+x+y'+z')(a'+x'+y+z) since
w+1=1 for every w in B.
x y z | f 0 0 0 | 0 0 0 1 | 0 0 1 1 | 0 0 1 0 | - 1 1 0 | - 1 1 1 | - 1 0 1 | 1 1 0 0 | 1
f = L = xy' r=U' = (x+yz')' = x'(y'+z) = x'y'+x'z d = y(z' +x) = yz' + xy