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The answer given is YES. But how is the dependency C -> DE preserved in this decomposition?
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Yes its dependency preserving ,

from R1 we get  : A-> BCD and  C->D

       and R2 : D->E

now see , C->D amd D->E so we can say C->DE

So all dependency are preserved .
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for dependency C--->DE --------------- 1 we have dependecy D--->E, now replace E with D in --------------- 1, we get C-------->DD means C---------->D. hence C------->D preserve in R1(ABCD) ?? am i right???
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What has explained is pseudo transitivity rule , and yes you're right.
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