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.
F  {EI > C} does not logically imply EI > C
since the closure of EI under F  {EI > C} is EI^+ = EI
and C is not a subset of EI.
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 Fall '10
 Dr.Butz
 Trigraph, Initialisms, Sue Grafton

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