I need help with this question,
Express the shaded portion of the Venn diagram in terms of unions, intersections or complements of sets.

enter image description here

I have done it as A Intersection B + A Intersection C - A Intersection B Intersection C + D Intersection E + (E-C)
i.e. $A\cap B \cup A\cap C - A\cap B\cap C \cup D \cap E \cup (E-C)$
I am not sure about the answer. Can someone please help me out with it? Thanks

  • $\begingroup$ $(A\cap B\cap C)$ is counted in $(A\cap B) + (A\cap C) $ twice, so if you subtract $(A\cap B\cap C)$, you're still left with one. $\endgroup$ – user13888 Nov 2 '11 at 21:16
  • $\begingroup$ @MikeWierzbicki: I just missed that. Thanks $\endgroup$ – user2857 Nov 2 '11 at 21:17
  • $\begingroup$ Also be careful with what you're taking out when you add $(E-C)$. $\endgroup$ – user13888 Nov 2 '11 at 21:17
  • 1
    $\begingroup$ @Akito: I've added TeX-ed version $A\cap B \cup A\cap C - A\cap B\cap C \cup D \cap E \cup (E-C)$ to make post more readable. Maybe you might have a look if this captures your intended meaning. You might also think about adding some parenthesis there, which might improve readability too. (I've tried to stick to your original text, just replaced $+$ by $\cup$ and intersection by $\cap$. $\endgroup$ – Martin Sleziak Nov 11 '11 at 8:46

You are asked to express the sets in terms of union, intersection, complements. In particular, the "minus" in your proposed solution would have to be expressed in terms of these. But aside from that minor comment, your answer is correct.

We do the problem in another way. We look at the four shaded parts, one after the other, going roughly downwards.
The top shaded part is inside $A$, inside $B$, outside $C$, outside $D$, and outside $E$. In symbols, it is $$A\cap B\cap C^c\cap D^c\cap E^c,$$ ($S^c$ denotes the complement of $S$).

The next one down is very similar, except we are outside $B$, inside $C$, with the rest exactly as above:
$$A\cap B^c\cap C\cap D^c\cap E^c.$$ The next one down is outside $A$, outside $B$, and inside $C$, $D$, and $E$: $$A^c\cap B^c\cap C\cap D\cap E.$$ Finally, the one at the bottom is inside $E$, and outside all the others: $$A^c\cap B^c\cap C^c\cap D^c\cap E.$$

Finally, we take the union of our four pieces. It is long, so we use a smaller font: $$\small(A\cap B\cap C^c\cap D^c\cap E^c)\cup(A\cap B^c\cap C\cap D^c\cap E^c)\cup(A^c\cap B^c\cap C\cap D\cap E)\cup(A^c\cap B^c\cap C^c\cap D^c\cap E).$$

Comment: This was a bit lengthy, but utterly mechanical. There are many other forms for the answer. For the purpose of circuit design, we would almost certainly want an expression that used fewer symbols. (We used a total of $19$ $\cap$ and/or $\cup$. The complement operations were not counted, because in circuit design they are often assumed to be "for free.")

  • $\begingroup$ $(A\cap B\cap C^c)\cup(A\cap B^c\cap C)\cup( D\cap E)\cup( C^c\cap E)$ would be shorter though not as systematic $\endgroup$ – Henry Nov 2 '11 at 21:48
  • $\begingroup$ @Henry: Thank you, the comment will be helpful to the OP. I wanted to give a procedure that was as mechanical as possible. $\endgroup$ – André Nicolas Nov 2 '11 at 22:13
  • $\begingroup$ Thankyou very good answer. $\endgroup$ – user2857 Nov 3 '11 at 5:53

Note that there are several ways to express the shaded area. Let us go through the parts:

  • There's a part of $A\cap B$, and a part of $A\cap C$. What missing from both is $A\cap B\cap C$. So we can write this part as $A\cap(B\triangle C)$, where $B\triangle C$ is the symmetric difference, that is $(B\cup C)\setminus(B\cap C)$.

  • On the bottom part there is a part of $E$ which meets $D$, and the part of $E$ which is not in $C$. We can write this as $(E\cap D)\cup(E\setminus C)$

Together we can write it all as: $$(A\cap(B\triangle C))\cup(E\cap D)\cup(E\setminus C)$$

  • $\begingroup$ I also wanted to ask if an area is not shaded, can we subtract it twice or more times? Does it matter? $\endgroup$ – user2857 Nov 2 '11 at 21:19
  • $\begingroup$ @Akito: The definition is: $$A\setminus B=\{x\mid x\in A\land x\notin B\}$$ Now suppose what is $(A\setminus B)\setminus B$? These are all the elements that are in $A\setminus B$ and not in $B$. However all the elements of $A\setminus B$ are not in $B$ by definition. So it does not matter to subtract twice. $\endgroup$ – Asaf Karagila Nov 2 '11 at 21:58

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