Proof on sets $(A\cap B)\cup(A\cap \bar{B}) = A$

Question

Original question :

To Prove : $(A\cap B)\cup(A\cap \bar{B}) = A$

My Response to it :

We have,

$(A\cap B)\cup(A\cap \bar{B}) = A$

$\Rightarrow (A \cap B) + (A \cap \bar{B}) - (A \cap B \cap A \cap \bar{B})$

$\Rightarrow (A \cap B) + (A \cap \bar {B}) - \emptyset$

$\Rightarrow (A \cap B) + (A \cap \bar {B})$

$\Rightarrow (A \cap B) + A \cap (U-B)$

$\Rightarrow (A \cap B) + (A \cap U) - (A \cap B)$

$\Rightarrow A$

The identity I used above was: $(X \cup Y) = X + Y - (X \cap Y)$

However, my professor says the identity I used is flawed and is wrong thus making my solution wrong too. Please provide your insight on what you think.

If anyone else is wondering, $X + Y$ is the symmetric difference; I don't know what the $-$ means though. — IAmNoOne, Apr 11 '15 at 01:54
As in the link above, by distributivity, $$(A\cap B)\cup(A\cap\overline{B})=A\cap(B\cup\overline{B})=A.$$ — Daniel W. Farlow, Apr 11 '15 at 01:59

score 1 · Answer 1 · answered Apr 11 '15 at 02:14

1

$(A \cap B) \cup (A \cap \bar B) =$

$=A \cap (B \cup \bar B)=$ (inverse distributive rule of $\cap$)
$=A \cap S=$ (S is the universal set)
$=A$ (q.e.d.)

answered Apr 11 '15 at 02:14

yellow_submarine

568

Proof on sets $(A\cap B)\cup(A\cap \bar{B}) = A$

1 Answers1