here if you Look carefully .....
common attributes in table (ABC) and (ACD)....is a set of attributes ....
so common set is AC....
now this set must be key in either of table ....either in ABC or in ACD....
but as per FD it totally fails ...because AC togather is not key in either of table ....so its lossy decomposition......
while in all other attributes ...a common attributes among any 2 table is a KEY ...but it fails in this decomposition ...so it is answer...
