1) R1(ABCD) AND R2(DE)
$(R1\cap R2)^{+}=R1 or R2$.THEN IT IS LOSSLESS JOIN DEPENDENCY.
HERE $(R1\cap R2)^{+}=DE$.// SO IT IS LOSSLESS JOIN DEPENDENCY.
2)DEPENDECY PRESERVING
R1(ABCD)---- A->BC,C->D
R2(DE)---- D->E.
G={A->BC,C->D,D->E}
by transtivity rule C->D AND D->E THEN C->E.
SO G={A->BC,C->DE,D->E}
SO IT IS DEPENDENCY PRESERVING