R1(A, B, C) and R2(C, D)
Let us find closure of F1 and F2
To find closure of F1, consider all combination of ABC. i.e., find closure of A, B, C, AB, BC and AC
Note ABC is not considered as it is always ABC
(A) += { A } // Trivial
(B)+ = { B } // Trivial
(C)+ = {C, A, D} //but D can't be in closure as D is not present R1.
= {C, A}
C--> A // Removing C from right side as it is trivial attribute
(AB)+ = {A, B, C, D}
= {A, B, C}
AB --> C // Removing AB from right side as these are trivial attributes
(BC)+ = {B, C, D, A}
= {A, B, C}
BC --> A // Removing BC from right side as these are trivial attributes
(AC)+ = {A, C, D}
NULL SET
F1 {C--> A, AB --> C, BC --> A}.
Similarly F2 { C--> D }
In the original Relation Dependency { AB --> C , C --> D , D --> A}.
AB --> C is present in F1.
C --> D is present in F2.
D --> A is not preserved.
F1 U F2 is a subset of F. So given decomposition is not dependency preserving.