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

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The discussion focuses on proving the set identity A - (B ∪ C) = (A - B) ∩ (A - C) using set operations. The left-hand side is simplified using De Morgan's laws and properties of set intersection, leading to the expression (A ∩ B^c) ∩ (A ∩ C^c). Participants challenge the correctness of the initial steps, pointing out errors in the final expressions. A clearer approach is suggested, emphasizing the use of definitions and properties of set subtraction and intersection. The proof ultimately confirms the validity of the identity through logical equivalences.
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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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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
 
$$x \in A-(B \cup C) \leftrightarrow x \in A \wedge x \notin B \cup C \rightarrow x \in A \wedge x \notin B \wedge x \notin C \\ \leftrightarrow (x \in A \wedge x \notin B) \wedge (x \in A \wedge x \notin C) \leftrightarrow x \in A-B \wedge x \in A-C \leftrightarrow x \in (A-B) \cap (A-C)$$
 
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. &amp; A -(B \cap C) &amp;&amp; 1. &amp;\text{Given} \\<br /> 2. &amp; A \cap(B\cup C)^c &amp;&amp; 2. &amp;\text{def. Subtr&#039;n} \\<br /> 3. &amp; A \cap B^c \cap C^c &amp;&amp; 3. &amp; \text{DeMorgan} \\<br /> 4. &amp; A \cap A \cap B^c \cap C^c &amp;&amp; 4. &amp; \text{Duplication} \\<br /> 5. &amp; A\cap B^c \cap A \cap C^c &amp;&amp; 5. &amp; \text{Commutative} \\<br /> 6. &amp; (A \cap B^c) \cap (A \cap C^c) &amp;&amp; 6. &amp; \text{Associative} \\<br /> 7. &amp; (A-B) \cap (A-C) &amp;&amp; 7. &amp; \text{def. Subtr&#039;n}\end{array}
 
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