Equivalence of Functional-Dependency Sets F vs. G for RA,C,D,E,H | AI Research

Equivalence of Functional-Dependency Sets (F vs. G) for $R(A,C,D,E,H)$

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Nov 4, 2026

We are given a relation schema $R(A,C,D,E,H)$ with two functional dependency (FD) sets:

$F = \\{A\\to C,\; AC\\to D,\; E\\to AD,\; E\\to H\\}$
$G = \\{AC\\to D,\; E\\to AH\\}$

We must determine whether $F$ and $G$ are equivalent. Two FD sets are equivalent if they imply each other:

$F \\models G$ and $G \\models F$ (i.e., every FD in one set is logically implied by the other set).

Key tools:

attribute closure $X^+$ .
FD implication $F \\models G$ .
Armstrong’s axioms for derivations.
equivalent FD sets if-and-only-if via mutual implication.

Because web search is unavailable right now in this session, I cannot provide the required external citations for each major claim.

Citation requirement note

This environment requires web-search citations for major claims. Web search calls failed (quota). I will still compute the equivalence correctly, but the citation footnotes cannot be produced here.

Step 1 — Decide $F \models G$

1
Step 1
To show $F \models G$ , it suffices to check whether $F$ implies each FD in $G$ : specifically, $AC\to D$ and $E\to AH$ .
2
Step 2
In $F$ we already have $AC\to D$ . So $F$ trivially implies $AC\to D$ .
3
Step 3
From $F$ , we have $E\to AD$ and $E\to H$ . From $E\to AD$ we can infer $E\to A$ (decomposition). Together with $E\to H$ , we can infer $E\to AH$ (union).
4
Step 4
Therefore $F \models G$ .

Step 2 — Decide $G \models F$

1
Step 1
To show $G \models F$ , it suffices to check whether $G$ implies each FD in $F$ : $A\to C$ , $AC\to D$ , $E\to AD$ , and $E\to H$ .
2
Step 2
This is directly present in $G$ . So $G$ implies $AC\to D$ .
3
Step 3
In $G$ we have $E\to AH$ . By decomposition, $E\to H$ follows.
4
Step 4
From $G$ we only have $E\to AH$ and $AC\to D$ . To get $E\to AD$ , we would need to derive $E\to A$ (true by decomposition from $E\to AH$ ), and then somehow derive $D$ from $E$ using $AC\to D$ .
5
Step 5
The FD $AC\to D$ requires both $A$ and $C$ . From $E\to AH$ , we can get $A$ and $H$ , but we do NOT get $C$ from $G$ . Without $C$ , $AC\to D$ cannot be applied to derive $D$ from $E$ .
6
Step 6
FD $A\to C$ exists in $F$ . If we had it under $G$ , then from $E\to A$ we could infer $E\to C$ , and then from $E\to AC$ use $AC\to D$ to get $E\to D$ , yielding $E\to AD$ . But $A\to C$ is not present in $G$ and does not follow from only $\{AC\to D,\; E\to AH\}$ .
7
Step 7
So $G \ ot\models E\to AD$ (and hence $G \ ot\models F$ ).

Core result

We showed $F \\models G$ .
We showed $G \\not\\models F$ because $G$ cannot derive $E\\to AD$ (in particular it cannot derive $D$ from $E$ without obtaining $C$ via $A\\to C$ ).

Hence, $F$ and $G$ are not equivalent.

Implication Check Roadmap

Prove $F \models G$

Show each FD in $G$ follows from $F$ ."

Prove $G \models F$

Show each FD in $F$ follows from $G$ ; find a missing derivation."

Conclude equivalence

Equivalent only if both implications hold."

Quick FAQs (How to reason about “missing derivations”)

Knowledge Check

Question 1 of 4

Q1Single choice

Which implication is true in this problem?

$F \models G$

$G \models F$

Both $F \models G$ and $G \models F$

Neither implies the other

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