Equivalence Of Fd Sets - Learning Module

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Testing Equivalence

The Equivalence Verification Problem

We have established what equivalence means (F ≡ G iff F⁺ = G⁺) and how to derive FDs. Now we address the practical challenge: Given two concrete FD sets F and G, how do we efficiently and correctly determine if they are equivalent?

This is not merely an academic exercise. Testing equivalence is required in numerous real-world scenarios:

Schema migration: Verify that a new database design preserves all constraints
Canonical cover computation: Confirm that the reduced FD set is equivalent to the original
Documentation reconciliation: Check if independently documented FD sets describe the same constraints
Decomposition verification: Ensure decomposed relations collectively preserve original dependencies
Tool validation: Verify that automated FD discovery tools produce correct results

What You Will Learn

By the end of this page, you will master the standard equivalence testing algorithm, understand why naive approaches fail, be able to implement efficient equivalence checks, handle edge cases and special scenarios, and apply these techniques to real database design problems.

The Naive Approach and Its Limitations

The most straightforward interpretation of the definition suggests: compute F⁺, compute G⁺, and check if they're equal. This approach is theoretically correct but practically impossible.

Why Full Closure Computation Fails

The closure F⁺ of an FD set can be astronomically large. For a relation with n attributes:

There are 2ⁿ possible subsets on the LHS
For each LHS, there are 2ⁿ possible subsets on the RHS
Total possible FDs: 2ⁿ × 2ⁿ = 2²ⁿ (ignoring the RHS ⊆ U constraint makes this even larger)
Even with 10 attributes: 2²⁰ ≈ 1 million FDs
With 20 attributes: 2⁴⁰ ≈ 1 trillion FDs

Exponential Explosion

Full closure computation is exponential in the number of attributes. For any practical schema, this approach is computationally infeasible. We need a smarter algorithm that avoids enumerating the full closure.

The Key Insight

We don't need to compute the full closure. Instead, we exploit the mutual coverage characterization:

$$F \equiv G \iff (F\ \text{covers}\ G) \land (G\ \text{covers}\ F)$$

To test if F covers G, we only need to check that each FD in G is derivable from F. We never enumerate FDs not in G.

Similarly for G covering F.

This reduces the problem from exponential (enumerate all FDs) to polynomial (check each FD in the given sets).

The Standard Equivalence Test Algorithm

The efficient algorithm for testing F ≡ G uses attribute closure to verify mutual coverage.

Algorithm: EQUIVALENT(F, G)

Equivalence Testing Algorithm
Pseudocode
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function EQUIVALENT(F, G):
    """
    Tests if F ≡ G (F and G have the same closure).
    Returns true if equivalent, false otherwise.
    
    Time Complexity: O((|F| + |G|) × max(|F|, |G|) × |attributes|)
    """
    
    # Step 1: Check if F covers G
    for each (X → Y) in G:
        closure_X_under_F = ATTRIBUTE_CLOSURE(X, F)
        if Y ⊄ closure_X_under_F:
            return false  # G has an FD not derivable from F
    
    # Step 2: Check if G covers F
    for each (X → Y) in F:
        closure_X_under_G = ATTRIBUTE_CLOSURE(X, G)
        if Y ⊄ closure_X_under_G:
            return false  # F has an FD not derivable from G
    
    # Both directions verified
    return true
 
 
function ATTRIBUTE_CLOSURE(X, FD_set):
    """
    Computes X⁺ under the given FD set.
    Standard fixed-point algorithm.
    """
    closure = X
    changed = true
    
    while changed:
        changed = false
        for each (A → B) in FD_set:
            if A ⊆ closure and B ⊄ closure:
                closure = closure ∪ B
                changed = true
    
    return closure

Correctness Proof

If the algorithm returns true: Every FD in G is in F⁺ (Step 1 passed), and every FD in F is in G⁺ (Step 2 passed). Hence F covers G and G covers F, so F ≡ G.
If F ≡ G: Then F⁺ = G⁺, so every FD in G is in G⁺ = F⁺ (checked in Step 1), and every FD in F is in F⁺ = G⁺ (checked in Step 2). The algorithm returns true.
Contrapositive: If the algorithm returns false, then either some FD in G is not in F⁺, or some FD in F is not in G⁺. Either way, F⁺ ≠ G⁺, so F ≢ G.

Complexity Analysis

Let:

|F| = n₁ (number of FDs in F)
|G| = n₂ (number of FDs in G)
|U| = k (number of attributes)
Step 1: n₂ calls to ATTRIBUTE_CLOSURE, each O(n₁ × k)
Step 2: n₁ calls to ATTRIBUTE_CLOSURE, each O(n₂ × k)
Total: O((n₁ × n₂ + n₂ × n₁) × k) = O(n₁ × n₂ × k)

This is polynomial—specifically, low-degree polynomial—making the algorithm highly practical.

Complete Worked Example

Let's walk through a complete equivalence test with detailed computations.

Problem:

Let U = {A, B, C, D} with:

F = {A → B, B → C, A → C, C → D}
G = {A → B, B → CD}

Determine if F ≡ G.

Step 1: Check if F covers G

For each FD in G, verify it's derivable from F.

FD: A → B

Compute {A}⁺_F:
- Start: {A}
- Apply A → B: {A, B}
- Apply B → C: {A, B, C}
- Apply A → C: (already have C)
- Apply C → D: {A, B, C, D}
Is B ⊆ {A, B, C, D}? Yes ✓

FD: B → CD

Compute {B}⁺_F:
- Start: {B}
- Apply A → B: (A not in closure, skip)
- Apply B → C: {B, C}
- Apply A → C: (A not in closure, skip)
- Apply C → D: {B, C, D}
Is CD ⊆ {B, C, D}? Yes ✓

Result: F covers G ✓

Step 2: Check if G covers F

For each FD in F, verify it's derivable from G.

FD: A → B

Compute {A}⁺_G:
- Start: {A}
- Apply A → B: {A, B}
- Apply B → CD: {A, B, C, D}
Is B ⊆ {A, B, C, D}? Yes ✓

FD: B → C

Compute {B}⁺_G:
- Start: {B}
- Apply A → B: (A not in closure, skip)
- Apply B → CD: {B, C, D}
Is C ⊆ {B, C, D}? Yes ✓

FD: A → C

Compute {A}⁺_G: (already computed above)
{A}⁺_G = {A, B, C, D}
Is C ⊆ {A, B, C, D}? Yes ✓

FD: C → D

Compute {C}⁺_G:
- Start: {C}
- Apply A → B: (A not in closure, skip)
- Apply B → CD: (B not in closure, skip)
Final: {C}
Is D ⊆ {C}? No ✗

Result: G does NOT cover F ✗

Conclusion: F ≢ G

The FD C → D is in F but cannot be derived from G. Therefore F⁺ ≠ G⁺, and the sets are NOT equivalent. F is 'stronger' than G—it imposes an additional constraint (C → D) that G doesn't capture.

Interpretation:

The sets F and G share some constraints but differ in an important way:

Both enforce: A → B, and transitively A → C, A → D
Both enforce: B → C, B → D
Only F enforces: C → D

In G, knowing C alone doesn't tell you D. You need B (which determines C and D together). In F, even knowing just C is enough to determine D.

This is a real semantic difference—databases designed with F vs. G behave differently for certain updates.

Edge Cases and Special Scenarios

The equivalence testing algorithm handles several special cases correctly. Understanding these ensures you don't make mistakes in edge scenarios.

Case 1: Empty FD Sets

Empty set F = {} has closure F⁺ containing only trivial FDs
{} ≡ {} is true
{} ≢ {A → B} for any non-trivial FD

The algorithm handles this: Step 1 vacuously passes (no FDs in G to check), Step 2 fails immediately on the first non-trivial FD in F.

Case 2: Identical FD Sets

If F = G, then trivially F ≡ G. The algorithm confirms this: every FD checked is in the same set it's being derived from.

Case 3: Subset Relationship

If F ⊂ G (F is a proper subset of G), they might or might not be equivalent:

If all extra FDs in G are derivable from F: F ≡ G
If some extra FD in G is not derivable from F: F ≢ G (but F covers G by subset property)

Wait—actually if F ⊂ G, then G covers F (since every FD in F is in G and thus in G⁺). But does F cover G?

If F covers G too, then F ≡ G despite F ⊂ G.

This happens when G contains redundant FDs.

Case 4: Trivial FDs Only

An FD set containing only trivial FDs (like {A → A, AB → A, AB → B}) is equivalent to the empty set. Both have closures containing exactly the trivial FDs.

Case 5: Different Attribute Sets

Strictly speaking, equivalence is defined for FD sets over the same attribute set U. If F and G mention different attributes, we extend each set's attribute universe to the union before comparing. FDs involving 'new' attributes are handled normally.

Case 6: Composite vs. Singleton RHS

As established earlier, {A → BC} ≡ {A → B, A → C}. The algorithm correctly verifies this:

A → BC is derivable from {A → B, A → C}: compute {A}⁺ = {A, B, C}, BC ⊆ {A, B, C} ✓
A → B derivable from {A → BC}: compute {A}⁺ = {A, B, C}, B ⊆ {A, B, C} ✓
A → C derivable from {A → BC}: same as above ✓

Practical Tip

Before running the full equivalence test, quickly check if one set is much smaller than the other. If |F| << |G|, consider whether all extra FDs in G could be derived from F. A quick sanity check can save computation time.

Implementation Considerations

When implementing equivalence testing in real software, several practical considerations arise.

Data Structure Choice

Representing FD sets efficiently matters:

Attributes as integers: Map attribute names to consecutive integers 0..n-1
Attribute sets as bitmasks: Subset operations become O(1) bit operations
FD as (LHS bitmask, RHS bitmask): Compact representation
FD set as sorted array: Fast iteration, easy deduplication

Efficient Implementation
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from typing import List, Set, Tuple
 
# FD represented as (LHS: frozenset, RHS: frozenset)
FD = Tuple[frozenset, frozenset]
 
def attribute_closure(x: frozenset, fds: List[FD]) -> frozenset:
    """Compute X⁺ under the given FD set."""
    closure = set(x)
    changed = True
    
    while changed:
        changed = False
        for lhs, rhs in fds:
            if lhs <= closure and not rhs <= closure:
                closure |= rhs
                changed = True
    
    return frozenset(closure)
 
def covers(f: List[FD], g: List[FD]) -> bool:
    """Check if F covers G (every FD in G derivable from F)."""
    for lhs, rhs in g:
        closure = attribute_closure(lhs, f)
        if not rhs <= closure:
            return False
    return True
 
def equivalent(f: List[FD], g: List[FD]) -> bool:
    """Check if F ≡ G (mutual coverage)."""
    return covers(f, g) and covers(g, f)
 
 
# Example usage
F = [
    (frozenset({'A'}), frozenset({'B'})),
    (frozenset({'B'}), frozenset({'C'})),
]
G = [
    (frozenset({'A'}), frozenset({'B'})),
    (frozenset({'B'}), frozenset({'C'})),
    (frozenset({'A'}), frozenset({'C'})),
]
 
print(f"F ≡ G: {equivalent(F, G)}")  # True

Optimization Opportunities

Early termination: Stop as soon as one FD fails the coverage check
Order FDs strategically: Check FDs with larger RHS first—they're more likely to fail coverage
Cache closures: If testing equivalence repeatedly (e.g., iterating toward canonical cover), cache computed closures
Parallel checking: The coverage checks for different FDs are independent; parallelize across FDs
Incremental testing: If comparing F to a modified version F', only re-check affected FDs

Bitmask Performance

For schemas with ≤64 attributes, representing attribute sets as 64-bit integers enables O(1) subset tests and set operations. This can speed up closure computation by an order of magnitude compared to set data structures.

Equivalence Testing in Practice

Let's examine how equivalence testing appears in real database design workflows.

Scenario 1: Validating Canonical Cover Computation

After computing a canonical cover Fc from F, we must verify Fc ≡ F:

Fc = COMPUTE_CANONICAL_COVER(F)
assert EQUIVALENT(F, Fc), "Canonical cover computation failed!"

This is a critical correctness check. If the canonical cover algorithm has a bug, equivalence testing catches it.

Scenario 2: Schema Migration

When migrating from database D1 to D2, constraints may be expressed differently:

F1 = EXTRACT_FDS(D1.schema)
F2 = EXTRACT_FDS(D2.schema)

if not EQUIVALENT(F1, F2):
    print("Warning: Schema migration changes constraints!")
    print("FDs in F1 not derivable from F2:", FIND_UNCOVERED(F1, F2))
    print("FDs in F2 not derivable from F1:", FIND_UNCOVERED(F2, F1))

Scenario 3: Normalization Verification

After decomposing R into R1, R2, ..., Rn, verify dependency preservation:

F_original = FDs of R
F_projected = UNION(PROJECT(F_original, Ri.attributes) for each Ri)

if F_projected covers F_original:
    print("Decomposition preserves dependencies")
else:
    print("Decomposition DOES NOT preserve dependencies!")
    print("Lost FDs:", FIND_UNCOVERED(F_original, F_projected))

Note: For dependency preservation, we only need F_projected to cover F_original, not full equivalence. F_projected might have additional FDs that weren't in F_original.

Scenario 4: Automated FD Discovery Validation

Data profiling tools discover FDs from data. To validate:

F_documented = FDs from schema documentation
F_discovered = FDs from data profiling tool

if not EQUIVALENT(F_documented, F_discovered):
    print("Discrepancy between documented and actual constraints!")
    # Investigate: is documentation outdated, or is data violating constraints?

Diagnostic Information

When equivalence testing fails, report which FDs couldn't be covered in which direction. This diagnostic information guides investigation and repair. Simply returning 'not equivalent' without details is unhelpful.

Comparison of FD Testing Problems
Problem	Question	Method	Complexity
Membership	Is X → Y in F⁺?	Attribute closure	O(\|F\| × k)
Coverage	Does F cover G?	\|G\| membership tests	O(\|G\| × \|F\| × k)
Equivalence	Is F ≡ G?	2 coverage tests	O((\|F\|+\|G\|) × n × k)
Superkey test	Is X a superkey?	Check if X⁺ = U	O(\|F\| × k)
Candidate key test	Is X a candidate key?	Superkey + minimality	O(\|X\| × \|F\| × k)

Testing Equivalence

The Equivalence Verification Problem

This is not merely an academic exercise. Testing equivalence is required in numerous real-world scenarios:

Schema migration: Verify that a new database design preserves all constraints
Canonical cover computation: Confirm that the reduced FD set is equivalent to the original
Documentation reconciliation: Check if independently documented FD sets describe the same constraints
Decomposition verification: Ensure decomposed relations collectively preserve original dependencies
Tool validation: Verify that automated FD discovery tools produce correct results

What You Will Learn

The Naive Approach and Its Limitations

The most straightforward interpretation of the definition suggests: compute F⁺, compute G⁺, and check if they're equal. This approach is theoretically correct but practically impossible.

Why Full Closure Computation Fails

The closure F⁺ of an FD set can be astronomically large. For a relation with n attributes:

There are 2ⁿ possible subsets on the LHS
For each LHS, there are 2ⁿ possible subsets on the RHS
Total possible FDs: 2ⁿ × 2ⁿ = 2²ⁿ (ignoring the RHS ⊆ U constraint makes this even larger)
Even with 10 attributes: 2²⁰ ≈ 1 million FDs
With 20 attributes: 2⁴⁰ ≈ 1 trillion FDs

Exponential Explosion

The Key Insight

We don't need to compute the full closure. Instead, we exploit the mutual coverage characterization:

$$F \equiv G \iff (F\ \text{covers}\ G) \land (G\ \text{covers}\ F)$$

To test if F covers G, we only need to check that each FD in G is derivable from F. We never enumerate FDs not in G.

Similarly for G covering F.

This reduces the problem from exponential (enumerate all FDs) to polynomial (check each FD in the given sets).

The Standard Equivalence Test Algorithm

The efficient algorithm for testing F ≡ G uses attribute closure to verify mutual coverage.

Algorithm: EQUIVALENT(F, G)

Equivalence Testing Algorithm
Pseudocode
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function EQUIVALENT(F, G):
    """
    Tests if F ≡ G (F and G have the same closure).
    Returns true if equivalent, false otherwise.
    
    Time Complexity: O((|F| + |G|) × max(|F|, |G|) × |attributes|)
    """
    
    # Step 1: Check if F covers G
    for each (X → Y) in G:
        closure_X_under_F = ATTRIBUTE_CLOSURE(X, F)
        if Y ⊄ closure_X_under_F:
            return false  # G has an FD not derivable from F
    
    # Step 2: Check if G covers F
    for each (X → Y) in F:
        closure_X_under_G = ATTRIBUTE_CLOSURE(X, G)
        if Y ⊄ closure_X_under_G:
            return false  # F has an FD not derivable from G
    
    # Both directions verified
    return true
 
 
function ATTRIBUTE_CLOSURE(X, FD_set):
    """
    Computes X⁺ under the given FD set.
    Standard fixed-point algorithm.
    """
    closure = X
    changed = true
    
    while changed:
        changed = false
        for each (A → B) in FD_set:
            if A ⊆ closure and B ⊄ closure:
                closure = closure ∪ B
                changed = true
    
    return closure

Correctness Proof

If the algorithm returns true: Every FD in G is in F⁺ (Step 1 passed), and every FD in F is in G⁺ (Step 2 passed). Hence F covers G and G covers F, so F ≡ G.
If F ≡ G: Then F⁺ = G⁺, so every FD in G is in G⁺ = F⁺ (checked in Step 1), and every FD in F is in F⁺ = G⁺ (checked in Step 2). The algorithm returns true.
Contrapositive: If the algorithm returns false, then either some FD in G is not in F⁺, or some FD in F is not in G⁺. Either way, F⁺ ≠ G⁺, so F ≢ G.

Complexity Analysis

Let:

|F| = n₁ (number of FDs in F)
|G| = n₂ (number of FDs in G)
|U| = k (number of attributes)
Step 1: n₂ calls to ATTRIBUTE_CLOSURE, each O(n₁ × k)
Step 2: n₁ calls to ATTRIBUTE_CLOSURE, each O(n₂ × k)
Total: O((n₁ × n₂ + n₂ × n₁) × k) = O(n₁ × n₂ × k)

This is polynomial—specifically, low-degree polynomial—making the algorithm highly practical.

Complete Worked Example

Let's walk through a complete equivalence test with detailed computations.

Problem:

Let U = {A, B, C, D} with:

F = {A → B, B → C, A → C, C → D}
G = {A → B, B → CD}

Determine if F ≡ G.

Step 1: Check if F covers G

For each FD in G, verify it's derivable from F.

FD: A → B

Compute {A}⁺_F:
- Start: {A}
- Apply A → B: {A, B}
- Apply B → C: {A, B, C}
- Apply A → C: (already have C)
- Apply C → D: {A, B, C, D}
Is B ⊆ {A, B, C, D}? Yes ✓

FD: B → CD

Compute {B}⁺_F:
- Start: {B}
- Apply A → B: (A not in closure, skip)
- Apply B → C: {B, C}
- Apply A → C: (A not in closure, skip)
- Apply C → D: {B, C, D}
Is CD ⊆ {B, C, D}? Yes ✓

Result: F covers G ✓

Step 2: Check if G covers F

For each FD in F, verify it's derivable from G.

FD: A → B

Compute {A}⁺_G:
- Start: {A}
- Apply A → B: {A, B}
- Apply B → CD: {A, B, C, D}
Is B ⊆ {A, B, C, D}? Yes ✓

FD: B → C

Compute {B}⁺_G:
- Start: {B}
- Apply A → B: (A not in closure, skip)
- Apply B → CD: {B, C, D}
Is C ⊆ {B, C, D}? Yes ✓

FD: A → C

Compute {A}⁺_G: (already computed above)
{A}⁺_G = {A, B, C, D}
Is C ⊆ {A, B, C, D}? Yes ✓

FD: C → D

Compute {C}⁺_G:
- Start: {C}
- Apply A → B: (A not in closure, skip)
- Apply B → CD: (B not in closure, skip)
Final: {C}
Is D ⊆ {C}? No ✗

Result: G does NOT cover F ✗

Conclusion: F ≢ G

Interpretation:

The sets F and G share some constraints but differ in an important way:

Both enforce: A → B, and transitively A → C, A → D
Both enforce: B → C, B → D
Only F enforces: C → D

In G, knowing C alone doesn't tell you D. You need B (which determines C and D together). In F, even knowing just C is enough to determine D.

This is a real semantic difference—databases designed with F vs. G behave differently for certain updates.

Edge Cases and Special Scenarios

The equivalence testing algorithm handles several special cases correctly. Understanding these ensures you don't make mistakes in edge scenarios.

Case 1: Empty FD Sets

Empty set F = {} has closure F⁺ containing only trivial FDs
{} ≡ {} is true
{} ≢ {A → B} for any non-trivial FD

The algorithm handles this: Step 1 vacuously passes (no FDs in G to check), Step 2 fails immediately on the first non-trivial FD in F.

Case 2: Identical FD Sets

If F = G, then trivially F ≡ G. The algorithm confirms this: every FD checked is in the same set it's being derived from.

Case 3: Subset Relationship

If F ⊂ G (F is a proper subset of G), they might or might not be equivalent:

If all extra FDs in G are derivable from F: F ≡ G
If some extra FD in G is not derivable from F: F ≢ G (but F covers G by subset property)

Wait—actually if F ⊂ G, then G covers F (since every FD in F is in G and thus in G⁺). But does F cover G?

If F covers G too, then F ≡ G despite F ⊂ G.

This happens when G contains redundant FDs.

Case 4: Trivial FDs Only

An FD set containing only trivial FDs (like {A → A, AB → A, AB → B}) is equivalent to the empty set. Both have closures containing exactly the trivial FDs.

Case 5: Different Attribute Sets

Case 6: Composite vs. Singleton RHS

As established earlier, {A → BC} ≡ {A → B, A → C}. The algorithm correctly verifies this:

A → BC is derivable from {A → B, A → C}: compute {A}⁺ = {A, B, C}, BC ⊆ {A, B, C} ✓
A → B derivable from {A → BC}: compute {A}⁺ = {A, B, C}, B ⊆ {A, B, C} ✓
A → C derivable from {A → BC}: same as above ✓

Practical Tip

Implementation Considerations

When implementing equivalence testing in real software, several practical considerations arise.

Data Structure Choice

Representing FD sets efficiently matters:

Attributes as integers: Map attribute names to consecutive integers 0..n-1
Attribute sets as bitmasks: Subset operations become O(1) bit operations
FD as (LHS bitmask, RHS bitmask): Compact representation
FD set as sorted array: Fast iteration, easy deduplication

Efficient Implementation
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from typing import List, Set, Tuple
 
# FD represented as (LHS: frozenset, RHS: frozenset)
FD = Tuple[frozenset, frozenset]
 
def attribute_closure(x: frozenset, fds: List[FD]) -> frozenset:
    """Compute X⁺ under the given FD set."""
    closure = set(x)
    changed = True
    
    while changed:
        changed = False
        for lhs, rhs in fds:
            if lhs <= closure and not rhs <= closure:
                closure |= rhs
                changed = True
    
    return frozenset(closure)
 
def covers(f: List[FD], g: List[FD]) -> bool:
    """Check if F covers G (every FD in G derivable from F)."""
    for lhs, rhs in g:
        closure = attribute_closure(lhs, f)
        if not rhs <= closure:
            return False
    return True
 
def equivalent(f: List[FD], g: List[FD]) -> bool:
    """Check if F ≡ G (mutual coverage)."""
    return covers(f, g) and covers(g, f)
 
 
# Example usage
F = [
    (frozenset({'A'}), frozenset({'B'})),
    (frozenset({'B'}), frozenset({'C'})),
]
G = [
    (frozenset({'A'}), frozenset({'B'})),
    (frozenset({'B'}), frozenset({'C'})),
    (frozenset({'A'}), frozenset({'C'})),
]
 
print(f"F ≡ G: {equivalent(F, G)}")  # True

Optimization Opportunities

Early termination: Stop as soon as one FD fails the coverage check
Order FDs strategically: Check FDs with larger RHS first—they're more likely to fail coverage
Cache closures: If testing equivalence repeatedly (e.g., iterating toward canonical cover), cache computed closures
Parallel checking: The coverage checks for different FDs are independent; parallelize across FDs
Incremental testing: If comparing F to a modified version F', only re-check affected FDs

Bitmask Performance

Equivalence Testing in Practice

Let's examine how equivalence testing appears in real database design workflows.

Scenario 1: Validating Canonical Cover Computation

After computing a canonical cover Fc from F, we must verify Fc ≡ F:

Fc = COMPUTE_CANONICAL_COVER(F)
assert EQUIVALENT(F, Fc), "Canonical cover computation failed!"

This is a critical correctness check. If the canonical cover algorithm has a bug, equivalence testing catches it.

Scenario 2: Schema Migration

When migrating from database D1 to D2, constraints may be expressed differently:

F1 = EXTRACT_FDS(D1.schema)
F2 = EXTRACT_FDS(D2.schema)

if not EQUIVALENT(F1, F2):
    print("Warning: Schema migration changes constraints!")
    print("FDs in F1 not derivable from F2:", FIND_UNCOVERED(F1, F2))
    print("FDs in F2 not derivable from F1:", FIND_UNCOVERED(F2, F1))

Scenario 3: Normalization Verification

After decomposing R into R1, R2, ..., Rn, verify dependency preservation:

F_original = FDs of R
F_projected = UNION(PROJECT(F_original, Ri.attributes) for each Ri)

if F_projected covers F_original:
    print("Decomposition preserves dependencies")
else:
    print("Decomposition DOES NOT preserve dependencies!")
    print("Lost FDs:", FIND_UNCOVERED(F_original, F_projected))

Note: For dependency preservation, we only need F_projected to cover F_original, not full equivalence. F_projected might have additional FDs that weren't in F_original.

Scenario 4: Automated FD Discovery Validation

Data profiling tools discover FDs from data. To validate:

F_documented = FDs from schema documentation
F_discovered = FDs from data profiling tool

if not EQUIVALENT(F_documented, F_discovered):
    print("Discrepancy between documented and actual constraints!")
    # Investigate: is documentation outdated, or is data violating constraints?

Diagnostic Information