The version of the answers is non-redundancy

Exercise 3.2.2

(i) S(A,B,C,D) with FD's A->B,B->C,B->D.

a.

<latex>A-$>$B,A-$>$C,A-$>$D,B-$>$C,B-$>$D </latex>

b.

<latex>A </latex>

c.

<latex>AB,AC,AD,ABC,ACD,ABD,ABCD </latex>

(ii) T(A,B,C,D) with FD's AB->C,BC->D,CD->A,and AD->B.

a.

<latex>AB-$>$C,AB-$>$D,AD-$>$C,AD-$>$B,BC-$>$A,BC-$>$D,CD-$>$A,CD-$>$B </latex>

b.

<latex>AB,AD,BC,CD</latex>

c.

<latex>ABC,ABD,ACD,BCD,ABCD </latex>

(iii) U(A,B,C,D) with FD's A->B,B->C,C->D,D->A.

a.

<latex>A-$>$B,A-$>$C,A-$>$D,B-$>$C,B-$>$D,B-$>$A,\\C-$>$D,C-$>$B,C-$>$A,D-$>$A,D-$>$C,D-$>$B</latex>

b.

<latex>A,B,C,D </latex>

c.

<latex>AB,AC,AD,BC,BD,CD,ABC,ABD,ACD,BCD,ABCD </latex>

Exercise 3.2.10

b) A→D, BD→E, AC→E, and DE→B .

<latex>AC-$>$B</latex>

c) AB→D, AC→E, BC→D, D→A, and E→B.

<latex>AC-$>$B, BC-$>$A</latex>

d) A→B, B→C, C→D, D→E, and E→A.

<latex>A-$>$B, B-$>$C, C-$>$A</latex>