Math Problem Statement

Prove that A−(B∪C)=(A−B)∩(A−C),for all sets A,B,C.

Solution

To prove that:

A(BC)=(AB)(AC)A - (B \cup C) = (A - B) \cap (A - C)

for all sets A,B,CA, B, C, we use set notation and logical reasoning.

Step 1: Definition of Set Difference

The set difference AXA - X is defined as:

AX={xAxX}A - X = \{ x \in A \mid x \notin X \}

Using this, we rewrite both sides of the equation in terms of set membership.

Step 2: Prove A(BC)(AB)(AC)A - (B \cup C) \subseteq (A - B) \cap (A - C)

Let xA(BC)x \in A - (B \cup C). By definition of set difference:

xAandxBC.x \in A \quad \text{and} \quad x \notin B \cup C.

Since xBCx \notin B \cup C, it follows that:

xBandxC.x \notin B \quad \text{and} \quad x \notin C.

Thus, by the definition of set difference:

xABandxAC.x \in A - B \quad \text{and} \quad x \in A - C.

Since xx is in both ABA - B and ACA - C, it must also be in their intersection:

x(AB)(AC).x \in (A - B) \cap (A - C).

This proves that:

A(BC)(AB)(AC).A - (B \cup C) \subseteq (A - B) \cap (A - C).

Step 3: Prove (AB)(AC)A(BC)(A - B) \cap (A - C) \subseteq A - (B \cup C)

Let x(AB)(AC)x \in (A - B) \cap (A - C). This means:

xABandxAC.x \in A - B \quad \text{and} \quad x \in A - C.

By the definition of set difference, we know that:

xA,xB,andxC.x \in A, \quad x \notin B, \quad \text{and} \quad x \notin C.

Since xBx \notin B and xCx \notin C, it follows that xBCx \notin B \cup C, meaning:

xA(BC).x \in A - (B \cup C).

Thus, we have shown that:

(AB)(AC)A(BC).(A - B) \cap (A - C) \subseteq A - (B \cup C).

Step 4: Conclusion

Since we have proved both subset inclusions:

A(BC)(AB)(AC)A - (B \cup C) \subseteq (A - B) \cap (A - C)

and

(AB)(AC)A(BC),(A - B) \cap (A - C) \subseteq A - (B \cup C),

it follows that:

A(BC)=(AB)(AC).A - (B \cup C) = (A - B) \cap (A - C).

Thus, the given set identity is proven.


Would you like further details or have any questions? Here are some related questions to explore:

  1. How does this proof change if we consider the union (AB)(AC)(A - B) \cup (A - C)?
  2. Can we generalize this proof for more than two sets, i.e., A(B1B2Bn)A - (B_1 \cup B_2 \cup \dots \cup B_n)?
  3. What is the analogous property for set intersection instead of union?
  4. How does this relate to De Morgan's laws in set theory?
  5. Can this proof be visualized using a Venn diagram?

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Math Problem Analysis

Mathematical Concepts

Set Theory
Set Operations
Set Difference
Set Union
Set Intersection

Formulas

A - (B ∪ C) = (A - B) ∩ (A - C)

Theorems

Set Difference Definition
Subset Definition
Set Operations

Suitable Grade Level

Grades 10-12