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

[KOO]

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 ∩ C^c)
(A ∩ B^c) ∩ (A∩ C^c)
(AUB) ∩ (AUC)

R.H.S = (A-B) ∩ (A-C)
(A∩B^c) ∩ (A∩C^c)
(AUB) ∩ (AUC)

L.H.S = R.H.S

Is this correct?

 

[Amer]

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 ∩ C^c)
(A ∩ B^c) ∩ (A∩ C^c)
(AUB) ∩ (AUC)

R.H.S = (A-B) ∩ (A-C)
(A∩B^c) ∩ (A∩C^c)
(AUB) ∩ (AUC)

L.H.S = R.H.S

Is this correct?

(A ∩ B^c) ∩ (A∩ C^c) = (AUB) ∩ (AUC) , this is not correct you could use
A ∩ B^c = A - B , and A∩ C^c=A - C
In fact
(A ∩ B^c) = (A^cUB)^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]

$$x \in A-(B \cup C) \leftrightarrow x \in A \wedge x \notin B \cup C \rightarrow x \in A \wedge x \notin B \wedge x \notin C \\ \leftrightarrow (x \in A \wedge x \notin B) \wedge (x \in A \wedge x \notin C) \leftrightarrow x \in A-B \wedge x \in A-C \leftrightarrow x \in (A-B) \cap (A-C)$$

 

[soroban]

Hello, KOO!

We should work on one side of the equation.

Let $$A, B, C$$ be three sets.
Prove that:.$$A - (B \cup C) \:=\: (A-B) \cap (A-C)$$

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