Candidate Keys via Functional Dependencies Attribute Closure | AI Research

Candidate Keys via Functional Dependencies (Attribute Closure)

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Sep 12, 2026

We’re given a relation $R(A,B,C,D,E)$ with functional dependencies (FDs): $A \rightarrow BC$ , $CD \rightarrow E$ , and $B \rightarrow D$ . Our goal is to find candidate keys—minimal attribute sets whose attribute closure equals all attributes of the relation, i.e., $X^+ = \{A,B,C,D,E\}$ . 2

Key learning terms: attribute closure functional dependency candidate key minimality superkey

Attribute closure - Wikipedia - Definition and method for computing $X^+$ from FDs. ↩
Candidate key - Wikipedia - Defines candidate keys and their relation to closures/superkeys. ↩

Attribute Closure & Candidate Keys (Functional Dependencies)

Strategy

Compute $X^+$ for relevant starting sets $X$ using the FDs.
Identify which $X$ are superkeys ( $X^+ = \{A,B,C,D,E\}$ ).
Enforce minimality (remove any attribute and re-check; if closure still gives all attributes, it wasn’t minimal). 2

Below, we use $A^+$ , $B^+$ , $C^+$ , $D^+$ , and $E^+$ -style reasoning plus combinations, focusing on reachability via the dependency graph implied by the FDs.

Candidate key - Wikipedia - Defines candidate keys and their relation to closures/superkeys. ↩
Functional dependency - Wikipedia - Background on FDs used to infer attributes and compute closures. ↩

Compute Candidate Keys by Attribute Closure

1
Step 1
Use $A \rightarrow BC$ , $B \rightarrow D$ , and $CD \rightarrow E$ .
2
Step 2
From $A$ , derive $B$ and $C$ using $A \rightarrow BC$ .
3
Step 3
Once $B$ is in the closure, add $D$ using $B \rightarrow D$ .
4
Step 4
With both $C$ and $D$ available, derive $E$ using $CD \rightarrow E$ .
5
Step 5
Check if the closure equals all attributes: $obreak{A,B,C,D,E}$ .
6
Step 6
Try proper subsets of $\{A\}$ (none exist except empty set), so minimality holds if $A$ is a superkey.
7
Step 7
Compute closures for other singletons and then check key candidates formed by combining only what can reach all attributes via the given FDs; eliminate any non-minimal superkeys by checking their subsets.

1) Compute closures

Closure of $A$

Start: $A^+ = \{A\}$ .

Using $A \rightarrow BC$ : add $B$ and $C$
$A^+ = \{A,B,C\}$
Now $B \rightarrow D$ : add $D$
$A^+ = \{A,B,C,D\}$
Now $CD \rightarrow E$ : since $C$ and $D$ are present, add $E$
$A^+ = \{A,B,C,D,E\}$

So, $A$ is a superkey.

Minimality check for $A$

The only proper subset of $\{A\}$ is the empty set, which cannot derive any of $A,B,C,D,E$ from the given FDs. Therefore, $A$ is minimal, hence a candidate key.

Candidate key candidate: $\boxed{\{A\}}$

2) Rule out other candidate keys

To ensure there are no other candidate keys, we must see whether any other attribute set $X$ can derive $A$ (since every candidate key must determine all attributes including $A$ ). But note:

None of the FDs have $A$ on the right-hand side.
Therefore, no attribute other than $A$ can produce $A$ via these FDs.

Thus, any superkey $X$ must already contain $A$ . Otherwise, $A \notin X^+$ , so $X^+$ cannot equal all attributes.

So any candidate key must be of the form $X \supseteq \{A\}$ . But we already found that $\{A\}$ itself is a superkey, and it is minimal. Hence, no larger set can be a candidate key (they would fail minimality).

Therefore the only candidate key is: [ \boxed{{A}} ]

type="tip" title="Pro Tip: Candidate keys must “cover” the left side of missing attributes" content="If an attribute never appears on the RHS of any FD (here, $A$ never appears on any FD’s RHS), then any candidate key must include it, because you can’t derive it from other attributes."

type="warning" title="Avoid a common mistake: assuming multiple candidates exist without testing minimality" content="Even if a set becomes a superkey (its closure gives all attributes), it’s not a candidate key unless removing any attribute breaks the ability to derive all attributes."

From FDs to Candidate Keys

Compute closures

Step 1

Use repeated application of $X \rightarrow Y$ rules to build $X^+$ ."

Check superkey condition

Step 2

Verify $X^+ = \{A,B,C,D,E\}$ ."

Enforce minimality

Step 3

Confirm no proper subset of $X$ is also a superkey."

Eliminate impossible keys

Step 4

If $A$ cannot be derived from others, any candidate key must contain $A$ ."

Closure Reachability Summary (Key Seeds)

Whether the seed’s closure reaches all attributes $\{A,B,C,D,E\}$ .

Common Questions

Candidate Key Mastery (FDs)

1 / 4

Question · Term

What is attribute closure $X^+$?

Click to reveal

Answer · Definition

The set of all attributes that can be inferred from X using the given FDs (by repeated application).

Knowledge Check

Question 1 of 4

Q1Single choice

Which set is a candidate key for $R(A,B,C,D,E)$ given $A \rightarrow BC$ , $CD \rightarrow E$ , and $B \rightarrow D$ ?

{A}

{B}

{C}

{D}

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