Prove that A - (B U C) = (A - B) ∩ (A - C)

  • MHB
  • Thread starter KOO
  • Start date
In summary, we are given the sets A, B, and C and we need to prove that A-(B ∪ C) = (A-B) ∩ (A-C). Using the definition of set subtraction, we can rewrite the left-hand side as A ∩ (B ∪ C)^c. Then, we can apply DeMorgan's law to get A ∩ B^c ∩ C^c. By duplicating the set A and rearranging the terms, we get (A ∩ B^c) ∩ (A ∩ C^c), which is equivalent to (A-B) ∩ (A-C) by the definition of set subtraction. Therefore, the left-hand side is equal
  • #1
KOO
19
0
Let A, B, and C be three sets. Prove that A-(BUC) = (A-B) ∩ (A-C)

Solution)

L.H.S = A - (B U C)
A ∩ (B U C)c
A ∩ (B c ∩ Cc)
(A ∩ Bc) ∩ (A∩ Cc)
(AUB) ∩ (AUC)

R.H.S = (A-B) ∩ (A-C)
(A∩Bc) ∩ (A∩Cc)
(AUB) ∩ (AUC)

L.H.S = R.H.S

Is this correct?
 
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  • #2
Re: Prove that A - (BUC) = (A-B) ∩ (A-C)

KOO said:
Let A, B, and C be three sets. Prove that A-(BUC) = (A-B) ∩ (A-C)

Solution)

L.H.S = A - (B U C)
A ∩ (B U C)c
A ∩ (B c ∩ Cc)
(A ∩ Bc) ∩ (A∩ Cc)
(AUB) ∩ (AUC)

R.H.S = (A-B) ∩ (A-C)
(A∩Bc) ∩ (A∩Cc)
(AUB) ∩ (AUC)

L.H.S = R.H.S

Is this correct?

(A ∩ Bc) ∩ (A∩ Cc) = (AUB) ∩ (AUC) , this is not correct you could use
A ∩ Bc = A - B , and A∩ Cc=A - C
In fact
(A ∩ Bc) = (AcUB)c

The red lines are false are and they are not useful, you solved it but the last lines are not equal to the previous ones
 
  • #3
$$x \in A-(B \cup C) \leftrightarrow x \in A \wedge x \notin B \cup C \rightarrow x \in A \wedge x \notin B \wedge x \notin C \\ \leftrightarrow (x \in A \wedge x \notin B) \wedge (x \in A \wedge x \notin C) \leftrightarrow x \in A-B \wedge x \in A-C \leftrightarrow x \in (A-B) \cap (A-C)$$
 
  • #4
Hello, KOO!

We should work on one side of the equation.

Let [tex]A, B, C[/tex] be three sets.
Prove that:.[tex]A - (B \cup C) \:=\: (A-B) \cap (A-C)[/tex]

[tex]\begin{array}{cccccc}
1. & A -(B \cap C) && 1. &\text{Given} \\
2. & A \cap(B\cup C)^c && 2. &\text{def. Subtr'n} \\
3. & A \cap B^c \cap C^c && 3. & \text{DeMorgan} \\
4. & A \cap A \cap B^c \cap C^c && 4. & \text{Duplication} \\
5. & A\cap B^c \cap A \cap C^c && 5. & \text{Commutative} \\
6. & (A \cap B^c) \cap (A \cap C^c) && 6. & \text{Associative} \\
7. & (A-B) \cap (A-C) && 7. & \text{def. Subtr'n}\end{array}[/tex]
 
  • #5


Yes, this is correct. You have correctly used the distributive property of set operations to show that the left-hand side (L.H.S) and right-hand side (R.H.S) are equal. Therefore, you have proved that A - (B U C) = (A - B) ∩ (A - C).
 

FAQ: Prove that A - (B U C) = (A - B) ∩ (A - C)

What is the meaning of "A - (B U C) = (A - B) ∩ (A - C)"?

The equation means that the elements in set A that are not in the union of sets B and C are the same as the elements in set A that are not in set B combined with the elements in set A that are not in set C.

How do you prove that the equation is true?

To prove the equation, we need to show that both sets have the same elements. This can be done by showing that any element that is in one set is also in the other set, and vice versa.

Can you provide an example to illustrate this equation?

Let's say we have set A = {1, 2, 3, 4, 5}, set B = {2, 4, 6} and set C = {3, 5, 7}. The left side of the equation would be A - (B U C) = A - {2, 3, 4, 5, 6, 7} = {1}. The right side of the equation would be (A - B) ∩ (A - C) = {1, 3, 5} ∩ {1, 2, 4} = {1}. Therefore, both sides have the same elements, proving the equation to be true.

What is the significance of this equation in mathematical proofs?

This equation is often used in mathematical proofs to show that two sets are equal. It is also used in set theory to demonstrate the relationship between sets and how they intersect and complement each other.

How does this equation relate to the concepts of complement and intersection?

The equation shows that the complement of the union of two sets is the same as the intersection of the individual complements of those sets. This illustrates the relationship between complement and intersection in set theory.

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