<s>
In	O
the	O
relational	B-Architecture
data	I-Architecture
model	I-Architecture
a	O
superkey	B-Application
is	O
a	O
set	O
of	O
attributes	O
that	O
uniquely	O
identifies	O
each	O
tuple	B-Application
of	O
a	O
relation	B-Language
.	O
</s>
<s>
Because	O
superkey	B-Application
values	O
are	O
unique	O
,	O
tuples	B-Application
with	O
the	O
same	O
superkey	B-Application
value	O
must	O
also	O
have	O
the	O
same	O
non-key	O
attribute	O
values	O
.	O
</s>
<s>
That	O
is	O
,	O
non-key	O
attributes	O
are	O
functionally	B-Application
dependent	I-Application
on	O
the	O
superkey	B-Application
.	O
</s>
<s>
The	O
set	O
of	O
all	O
attributes	O
is	O
always	O
a	O
superkey	B-Application
(	O
the	O
trivial	O
superkey	B-Application
)	O
.	O
</s>
<s>
Tuples	B-Application
in	O
a	O
relation	B-Language
are	O
by	O
definition	O
unique	O
,	O
with	O
duplicates	O
removed	O
after	O
each	O
operation	O
,	O
so	O
the	O
set	O
of	O
all	O
attributes	O
is	O
always	O
uniquely	O
valued	O
for	O
every	O
tuple	B-Application
.	O
</s>
<s>
A	O
candidate	B-Application
key	I-Application
(	O
or	O
minimal	O
superkey	B-Application
)	O
is	O
a	O
superkey	B-Application
that	O
ca	O
n't	O
be	O
reduced	O
to	O
a	O
simpler	O
superkey	B-Application
by	O
removing	O
an	O
attribute	O
.	O
</s>
<s>
For	O
example	O
,	O
in	O
an	O
employee	O
schema	O
with	O
attributes	O
employeeID	O
,	O
name	O
,	O
job	O
,	O
and	O
departmentID	O
,	O
if	O
employeeID	O
values	O
are	O
unique	O
then	O
employeeID	O
combined	O
with	O
any	O
or	O
all	O
of	O
the	O
other	O
attributes	O
can	O
uniquely	O
identify	O
tuples	B-Application
in	O
the	O
table	O
.	O
</s>
<s>
Each	O
combination	O
,	O
{employeeID},	O
{	O
employeeID	O
,	O
name}	O
,	O
{	O
employeeID	O
,	O
name	O
,	O
job}	O
,	O
and	O
so	O
on	O
is	O
a	O
superkey	B-Application
.	O
</s>
<s>
 { employeeID } 	O
is	O
a	O
candidate	B-Application
key	I-Application
,	O
since	O
no	O
subset	O
of	O
its	O
attributes	O
is	O
also	O
a	O
superkey	B-Application
.	O
</s>
<s>
{	O
employeeID	O
,	O
name	O
,	O
job	O
,	O
departmentID}	O
is	O
the	O
trivial	O
superkey	B-Application
.	O
</s>
<s>
If	O
attribute	O
set	O
K	O
is	O
a	O
superkey	B-Application
of	O
relation	B-Language
R	O
,	O
then	O
at	O
all	O
times	O
it	O
is	O
the	O
case	O
that	O
the	O
projection	B-Algorithm
of	O
R	O
over	O
K	O
has	O
the	O
same	O
cardinality	B-Application
as	O
R	O
itself	O
.	O
</s>
<s>
Second	O
,	O
eliminate	O
all	O
the	O
sets	O
which	O
do	O
not	O
meet	O
superkey	B-Application
's	O
requirement	O
.	O
</s>
<s>
For	O
example	O
,	O
{	O
Monarch	O
Name	O
,	O
Royal	O
House}	O
cannot	O
be	O
a	O
superkey	B-Application
because	O
for	O
the	O
same	O
attribute	O
values	O
(	O
Edward	O
,	O
Plantagenet	O
)	O
,	O
there	O
are	O
two	O
distinct	O
tuples	B-Application
:	O
</s>
<s>
Finally	O
,	O
after	O
elimination	O
,	O
the	O
remaining	O
sets	O
of	O
attributes	O
are	O
the	O
only	O
possible	O
superkeys	B-Application
in	O
this	O
example	O
:	O
</s>
<s>
In	O
reality	O
,	O
superkeys	B-Application
cannot	O
be	O
determined	O
simply	O
by	O
examining	O
one	O
set	O
of	O
tuples	B-Application
in	O
a	O
relation	B-Language
.	O
</s>
<s>
A	O
superkey	B-Application
defines	O
a	O
functional	B-Application
dependency	I-Application
constraint	O
of	O
a	O
relation	B-Language
schema	I-Language
which	O
must	O
hold	O
for	O
all	O
possible	O
instance	O
relations	O
of	O
that	O
relation	B-Language
schema	I-Language
.	O
</s>
