October 2, 1995

				cs330 - Discrete Structures
				===========================
		          		Fall 1995
Hw #2 - Solutions
=================
Note: 	In the following solutions 
		V stands for the universal quantifier
		E stands for the existential quantifier
		^ stands for conjunction
		v stands for disjunction
		== means equal
		!= means not equal

	For the sake of readability, I've left a space between succesive
	quantifiers.


10/34
====
a)
Vx F(x, Jerry)

b)
Vx F(Evelyn, x) or
Vy F(Evelyn, y)		whichever you prefer

c)
Vx Ey F(x, y)

d)
~Ex Vy F(x, y)

e)
Vx Ey F(y, x)

f)
~Ex (F(x, Fred) ^ F(x, Jerry))

g)
Ex Ey((x!=y) ^ F(Nancy, x) ^ F(nancy, y) ^ Vz(F(Nancy, z) -> ((z==x) v (z==y))))

h)
Ex(VyF(y, x) ^ Vz(VyF(y, z) -> (z==x)))

i)
Vx ~F(x, x)

j)
Ex(F(x, x) ^ Ey(F(x, y) ^ Vz(F(x, z) -> (z==y))))


16/35
=====
L(x)       = "x likes mathematics"
S(x)       = "x has never seen a computer"
T(x, y)    = "x has taken the math class y at this school"
B(x, r)    = "x has been in r"
I(r, b)    = "r in b"

a)
x in the set {students in this class}
Vx L(x)
~(Vx L(x)) = Ex ~L(x)
"There is a student in this class who does not like mathematics"	or
"Some student in this class does not like mathematics"

b)
x in the set {students in this class}
Ex S(x)
~(Ex S(x)) = Vx ~S(x)
"every student in this class has seen a computer"

c)
x in the set {students in this class}
y in the set {math classes}
Ex VyT(x, y)
~(Ex VyT(x, y)) = Vx Ey ~T(x, y)
"every student in this class has not taken some math course"

d)
x in the set {students in this class}
r in the set {rooms}
b in the set {buildings on campus}
Ex Vb Er(I(r, b) ^ B(x, r))
~(Ex Vb Er(I(r, b) ^ B(x, r))) = Vx Eb Vr ~(I(r, b) ^ B(x, r))
			     <=> Vx Eb Vr (~I(r, b) v ~B(x, r))
"For every student in this class there exists a building such that for
every room, either that room is not in the building or the student has not
been in that room"


20/36
=====
x in the set {animals} for all parts of this problem
a)
Vx(P(x) -> ~S(x))

b)
~Ex(R(x) ^ ~S(x))

c)
Vx(Q(x) -> P(x))

d)
Vx (Q(x) -> ~R(x))


10/45
=====
a) 0	no elements in the empty set
b) 1	one element in the set, this is {}
c) 2	the elements are {} and {{}}
d) 3	the elements are {}, {{}} and {{},{{}}}


20/45
=====
a)
AxBxC = { (a,x,0), (a,x,1),
          (a,y,0), (a,y,1),
          (b,x,0), (b,x,1),
          (b,y,0), (b,y,1),
          (c,x,0), (c,x,1),
          (c,y,0), (c,y,1)
        }

b)
CxBxA = { (0,x,a), (0,x,b), (0,x,c),
          (0,y,a), (0,y,b), (0,y,c),
          (1,x,a), (1,x,b), (1,x,c),
          (1,y,a), (1,y,b), (1,y,c)
        }

c)
CxAxB = { (0,a,x), (0,a,y),
          (0,b,x), (0,b,y),
          (0,c,x), (0,c,y),
          (1,a,x), (1,a,y),
          (1,b,x), (1,b,y),
          (1,c,x), (1,c,y)
        }

d)
BxBxB = { (x,x,x), (x,x,y),
          (x,y,x), (x,y,y),
          (y,x,x), (y,x,y),
          (y,y,x), (y,y,y)
        }


6/70
====
a) injective	easy to check that for every x1!=x2 we have f(x1)!=f(x2)
b) not injective because f(a)=f(b)=b
c) not injective because f(a)=f(d)=d

8/70
====
a) injective
	assume f(x1) = f(x2) which means
               x1-1 = x2-1
               x1 = x2		which completes the proof.

b) not injective
	f(2)=f(-2)=5 yet 2 does not equal -2

c) injective
	f(x1) = f(x2)
        x1^3 = x2^3
        x1^3 - x2^3 = 0
        (x1 - x2)(x1^2 + x1*x2 + x2^2) = 0
	(x1 - x2) = 0	(the other paranthesis is always non-zero!)
	x1 = x2		which completes the proof

d) non-injective
	ceiling(3/2) = ceiling(4/2) = 2


12/70
=====
a) f(S) = {1}
b) f(S) = {-1, 1, 5, 9, 15}
c) f(S) = {0, 1, 2}
d) f(S) = {0, 1, 5, 16}