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

KOO

New member
Oct 19, 2013
19
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?
 

Amer

Active member
Mar 1, 2012
275
Re: Prove that A - (BUC) = (A-B) ∩ (A-C)

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
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,718
$$x \in A-(B \cup C) \leftrightarrow x \in A \wedge x \notin B \cup C \rightarrow x \in A \wedge x \notin B \wedge x \notin C \\ \leftrightarrow (x \in A \wedge x \notin B) \wedge (x \in A \wedge x \notin C) \leftrightarrow x \in A-B \wedge x \in A-C \leftrightarrow x \in (A-B) \cap (A-C)$$
 

soroban

Well-known member
Feb 2, 2012
409
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]