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Given $A,B,C\subseteq X$, prove $(A\cup B)\cap (\bar{A}\cup C) = (A\cap C)\cup (\bar{A}\cap B)$.

My attempt:

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

I'm stuck. I don't see what I did wrong, because $A\cap C$ isn't equal to $(A\cup B)\cap C$.

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    what does $\bar{A}$ mean? closure? – 2017-02-02
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    It's the complement of A. – 2017-02-02

1 Answers 1

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OK, from your last line:

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

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

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

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

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

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

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

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

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

For the last step: the two bigger terms get absorbed by the two smaller terms, since as a general Absorption principle you have:

$X \cup (X \cap Y) = X$

Also, For that step in the middle I use:

$X = X \cap (Y \cup \bar{Y})$

If you don't see how that is true, note that $Y \cup \bar{Y}$ has all elements, so the intersection between that and any set $X$ is $X$ itself.

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    I have corrected that error. I just wrote it wrong. – 2017-02-02
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    That means $(A\cup B)\cap C=A\cap C$? I tried proving that using Venn diagrams and I don't see that... – 2017-02-02
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    @Alberto $((A\cup B)\cap C)\cup (\bar{A}\cap B)=A\cap C \cup (\bar{A}\cap B)$ does not imply $(A\cup B)\cap C=A\cap C$, in general. – 2017-02-02
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    @user160738 But the way Bram has solve it do imply it as far as I see... he doesn't use the intersection of B and the complement of A – 2017-02-02
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    @Alberto He does use that at the last line; namely the fact that $(B\cap C\cap \bar{A})\cup (\bar{A}\cap B)=(\bar{A}\cap B)$. If that wasn't there, you could not simplify $(B\cap C\cap \bar{A})$ in such a way. So what his answer illustrates is that $(A\cup B)\cap C = (A\cap C)\cup (B \cap C \cap A) \cup (B \cap C \cap \bar{A} )$ – 2017-02-02
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    Yes, sorry... Now I see it. Thank you – 2017-02-02
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    @Alberto I added a few more steps to make things more clear. – 2017-02-02