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

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SUMMARY

The discussion centers on proving the set identity A - (B ∪ C) = (A - B) ∩ (A - C) using set theory principles. The proof employs De Morgan's laws and the definitions of set subtraction. Key steps include transforming the left-hand side (L.H.S) into an intersection of complements and demonstrating that it equals the right-hand side (R.H.S) through associative and commutative properties. The conclusion confirms the identity as valid.

PREREQUISITES
  • Understanding of set theory concepts, including set subtraction and union.
  • Familiarity with De Morgan's laws in set operations.
  • Knowledge of basic logical equivalences in mathematics.
  • Ability to manipulate set expressions using associative and commutative properties.
NEXT STEPS
  • Study De Morgan's laws in depth for better comprehension of set operations.
  • Explore advanced set theory concepts, such as Cartesian products and power sets.
  • Learn about logical equivalences and their applications in mathematical proofs.
  • Practice proving other set identities to enhance problem-solving skills in set theory.
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KOO
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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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