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26 votes
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Let $A$ and $B$ be sets and let $A^c$ and $B^c$ denote the complements of the sets $A$ and $B$. The set $(A-B) \cup (B-A) \cup (A \cap B)$ is equal to

  1. $A \cup B$

  2. $A^c \cup B^c$

  3. $A \cap B$

  4. $A^c \cap B^c$

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8 Answers

Best answer
28 votes
28 votes
$(A - B) \cup (B - A) \cup (A \cap B)$
$A - B$     is A but not B.  i.e. only A
$B - A$    is B but not A.  i.e. only B
$A \cap B$     is A and B both

Union of all is (only A) U (only B) U (both A and B)
$= A \cup B$

Correct Answer: $A$
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12 votes
12 votes

We can solve it using boolean algebra also

Given expression is :

(A−B)∪(B−A)∪(A∩B)

Which can be written in boolean algebra as

AB' + BA' + AB           { A-B can be expressed as A ∩B' }

Which gives A+B which is nothing but A U B.

6 votes
6 votes
Think like OSA:

This is like the equation of HALF ADDER!
(Symmetric difference= XOR)  U (AB)

=A XOR B + AB

AND HALF ADDER GIVES SUM.
SO A + B . which is A U B.
:)
Answer:

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