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

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Discussion Overview

The discussion revolves around proving the set identity A - (B ∪ C) = (A - B) ∩ (A - C). Participants explore various approaches to demonstrate this equality, focusing on set operations and properties.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Post 1 presents a solution using set operations but questions the correctness of the final steps.
  • Post 2 reiterates the solution from Post 1 and challenges the validity of the last lines, suggesting alternative expressions for A ∩ Bc and A ∩ Cc.
  • Post 3 provides a logical proof using element-wise reasoning to show the equivalence of the two sides of the equation.
  • Post 4 outlines a step-by-step approach to proving the identity, employing definitions and properties of set operations, but does not explicitly address the correctness of previous claims.

Areas of Agreement / Disagreement

Participants express differing views on the correctness of the solutions presented, particularly regarding the manipulation of set expressions. There is no consensus on the validity of the approaches or the final equality.

Contextual Notes

Some participants' arguments depend on specific interpretations of set operations and properties, which may not be universally accepted. The discussion includes unresolved steps and assumptions in the proofs presented.

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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