CS375 2010 Fall Solution Key #3
1.
Given F = {AB --> C, C --> B}, then
F^+ = {A --> A, B --> B, C --> B, C --> C, C --> BC,
AB --> A, AB --> B, AB --> C, AB --> AB, AB --> AC, AB --> BC, AB --> ABC,
AC --> A, AC --> B, AC --> C, AC --> AB, AC --> AC, AC --> BC, AC --> ABC,
BC --> B, BC --> C, BC --> BC, ABC --> A, ABC --> B, ABC --> C,
ABC --> AB, ABC --> AC, ABC --> BC, ABC --> ABC}
2. Let G = {A --> C, AB --> C, C --> DI, CD --> I, EC --> AB, EI --> C}.
Then NONREDUNDANT(G) sets
F = {A --> C, AB --> C, C --> DI, CD --> I, EC --> AB, EI --> C}.
Consider A --> C.
F - {A --> C} does not logically imply A --> C
since the closure of A under F - {A --> C} is A^+ = A
and C is not a subset of A.
Consider AB --> C.
F - {AB --> C} logically implies AB --> C
since the closure of AB under F - {AB --> C} is AB^+ = ABCDI
and C is a subset of ABCDI.
Thus, the FD AB --> C is removed and
F = {A --> C, C --> DI, CD --> I, EC --> AB, EI --> C}.
Consider C --> DI.
F - { C --> DI } does not logically imply C --> DI
since the closure of C under F - {C --> DI} is C^+ = C
and DI is not a subset of C.
Consider CD --> I.
F - {CD --> I} logically implies CD --> I
since the closure of CD under F - {CD --> I} is CD^+ = CDI
and I is a subset of CDI.
Thus, the FD CD --> I is removed and
F = {A --> C, C --> DI, EC --> AB, EI --> C}.
Consider EC --> AB.
F - {EC --> AB} does not logically imply EC --> AB
since the closure of EC under F - {EC --> AB} is EC^+ = ECDI
and AB is not a subset of ECDI.
Lastly, consider EI --> C.