Mixing
Prove $ Atriangle(Bbackslash C)=(Atriangle B)backslash C $ iff $Acap C=emptyset$
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Prove$$ Atriangle(Bbackslash C)=(Atriangle B)backslash C $$
If and only if $Acap C=emptyset$.
I should prove it using just or and and. Like this:
The first inclusion:
$$ x in Atriangle(Bbackslash C)\
[x in A backslash(B backslash C)] vee [xin(B backslash C) backslash A]$$
I think the next step is wrong:
$$
[x in (A backslash B) backslash C)] vee [xin(B backslash A) backslash C]\
[x in (A backslash B) vee xin(B backslash A) ] land xnotin C\
(Atriangle B)backslash C
$$
elementary-set-theory
asked Jan 5 at 1:50
Nemanja DjordjevicNemanja Djordjevic
888
$endgroup$
|
show 3 more comments
$begingroup$
Prove$$ Atriangle(Bbackslash C)=(Atriangle B)backslash C $$
If and only if $Acap C=emptyset$.
I should prove it using just or and and. Like this:
The first inclusion:
$$ x in Atriangle(Bbackslash C)\
[x in A backslash(B backslash C)] vee [xin(B backslash C) backslash A]$$
I think the next step is wrong:
$$
[x in (A backslash B) backslash C)] vee [xin(B backslash A) backslash C]\
[x in (A backslash B) vee xin(B backslash A) ] land xnotin C\
(Atriangle B)backslash C
$$
elementary-set-theory
asked Jan 5 at 1:50
Nemanja DjordjevicNemanja Djordjevic
888
$endgroup$
$begingroup$
Yes, that step is indeed wrong, because the set difference operation is not associative, so let's not do that. I'd suggest that you start by drawing Venn diagrams for both $Abigtriangleup(Bsetminus C)$ and $(Abigtriangleup B)setminus C$. You will see that, in general, they are not equal, as well as why $Acap C=varnothing$ is needed to make them equal. That may give you a hint how to approach the problem.
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– zipirovich
Jan 5 at 2:06
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I still don't know.
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– Nemanja Djordjevic
Jan 5 at 2:15
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Did you draw Venn diagrams for both expressions?
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– zipirovich
Jan 5 at 2:17
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yes and they are the same
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– Nemanja Djordjevic
Jan 5 at 2:35
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No, they are not.
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– zipirovich
Jan 5 at 2:36
|
show 3 more comments
$begingroup$
Prove$$ Atriangle(Bbackslash C)=(Atriangle B)backslash C $$
If and only if $Acap C=emptyset$.
I should prove it using just or and and. Like this:
The first inclusion:
$$ x in Atriangle(Bbackslash C)\
[x in A backslash(B backslash C)] vee [xin(B backslash C) backslash A]$$
I think the next step is wrong:
$$
[x in (A backslash B) backslash C)] vee [xin(B backslash A) backslash C]\
[x in (A backslash B) vee xin(B backslash A) ] land xnotin C\
(Atriangle B)backslash C
$$
elementary-set-theory
asked Jan 5 at 1:50
Nemanja DjordjevicNemanja Djordjevic
888
$endgroup$
Prove$$ Atriangle(Bbackslash C)=(Atriangle B)backslash C $$
If and only if $Acap C=emptyset$.
I should prove it using just or and and. Like this:
The first inclusion:
$$ x in Atriangle(Bbackslash C)\
[x in A backslash(B backslash C)] vee [xin(B backslash C) backslash A]$$
I think the next step is wrong:
$$
[x in (A backslash B) backslash C)] vee [xin(B backslash A) backslash C]\
[x in (A backslash B) vee xin(B backslash A) ] land xnotin C\
(Atriangle B)backslash C
$$
elementary-set-theory
elementary-set-theory
asked Jan 5 at 1:50
Nemanja DjordjevicNemanja Djordjevic
888
asked Jan 5 at 1:50
Nemanja DjordjevicNemanja Djordjevic
888
edited Jan 5 at 17:55
Nemanja Djordjevic
asked Jan 5 at 1:50
Nemanja DjordjevicNemanja Djordjevic
888
asked Jan 5 at 1:50
Nemanja DjordjevicNemanja Djordjevic
888
asked Jan 5 at 1:50
Nemanja DjordjevicNemanja Djordjevic
888
888
$begingroup$
Yes, that step is indeed wrong, because the set difference operation is not associative, so let's not do that. I'd suggest that you start by drawing Venn diagrams for both $Abigtriangleup(Bsetminus C)$ and $(Abigtriangleup B)setminus C$. You will see that, in general, they are not equal, as well as why $Acap C=varnothing$ is needed to make them equal. That may give you a hint how to approach the problem.
$endgroup$
– zipirovich
Jan 5 at 2:06
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I still don't know.
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– Nemanja Djordjevic
Jan 5 at 2:15
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Did you draw Venn diagrams for both expressions?
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– zipirovich
Jan 5 at 2:17
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yes and they are the same
$endgroup$
– Nemanja Djordjevic
Jan 5 at 2:35
$begingroup$
No, they are not.
$endgroup$
– zipirovich
Jan 5 at 2:36
|
show 3 more comments
$begingroup$
Yes, that step is indeed wrong, because the set difference operation is not associative, so let's not do that. I'd suggest that you start by drawing Venn diagrams for both $Abigtriangleup(Bsetminus C)$ and $(Abigtriangleup B)setminus C$. You will see that, in general, they are not equal, as well as why $Acap C=varnothing$ is needed to make them equal. That may give you a hint how to approach the problem.
$endgroup$
– zipirovich
Jan 5 at 2:06
$begingroup$
I still don't know.
$endgroup$
– Nemanja Djordjevic
Jan 5 at 2:15
$begingroup$
Did you draw Venn diagrams for both expressions?
$endgroup$
– zipirovich
Jan 5 at 2:17
$begingroup$
yes and they are the same
$endgroup$
– Nemanja Djordjevic
Jan 5 at 2:35
$begingroup$
No, they are not.
$endgroup$
– zipirovich
Jan 5 at 2:36
$begingroup$
Yes, that step is indeed wrong, because the set difference operation is not associative, so let's not do that. I'd suggest that you start by drawing Venn diagrams for both $Abigtriangleup(Bsetminus C)$ and $(Abigtriangleup B)setminus C$. You will see that, in general, they are not equal, as well as why $Acap C=varnothing$ is needed to make them equal. That may give you a hint how to approach the problem.
$endgroup$
– zipirovich
Jan 5 at 2:06
$begingroup$
Yes, that step is indeed wrong, because the set difference operation is not associative, so let's not do that. I'd suggest that you start by drawing Venn diagrams for both $Abigtriangleup(Bsetminus C)$ and $(Abigtriangleup B)setminus C$. You will see that, in general, they are not equal, as well as why $Acap C=varnothing$ is needed to make them equal. That may give you a hint how to approach the problem.
$endgroup$
– zipirovich
Jan 5 at 2:06
$begingroup$
I still don't know.
$endgroup$
– Nemanja Djordjevic
Jan 5 at 2:15
$begingroup$
I still don't know.
$endgroup$
– Nemanja Djordjevic
Jan 5 at 2:15
$begingroup$
Did you draw Venn diagrams for both expressions?
$endgroup$
– zipirovich
Jan 5 at 2:17
$begingroup$
Did you draw Venn diagrams for both expressions?
$endgroup$
– zipirovich
Jan 5 at 2:17
$begingroup$
yes and they are the same
$endgroup$
– Nemanja Djordjevic
Jan 5 at 2:35
$begingroup$
yes and they are the same
$endgroup$
– Nemanja Djordjevic
Jan 5 at 2:35
$begingroup$
No, they are not.
$endgroup$
– zipirovich
Jan 5 at 2:36
$begingroup$
No, they are not.
$endgroup$
– zipirovich
Jan 5 at 2:36
|
show 3 more comments
4 Answers
4
active
oldest
votes
$begingroup$
These are the Venn diagrams for both expressions. As we can see, they differ precisely by $Acap C$, which is why $Acap C=varnothing$ is required to make them equal. Also, notice that $(Abigtriangleup B)setminus Csubseteq Abigtriangleup(Bsetminus C)$ always; it's only the opposite inclusion that needs $Acap C=varnothing$.
If you're expected to prove the claim using just or and and logical connectors, then your first steps look good, but then I guess you should continue expanding all the definitions. Let's see how that looks for the inclusion $Abigtriangleup(Bsetminus C)subseteq(Abigtriangleup B)setminus C)$ that you started working on.
$$begin{split}
xin Abigtriangleup(Bsetminus C) &iff left[xin Asetminus(Bsetminus C)right]veeleft[xin(Bsetminus C)setminus Aright] \
&iff left[xin Awedge xnotin(Bsetminus C)right]veeleft[xin(Bsetminus C)wedge xnotin Aright] \
&iff left[xin Awedgenegleft(xin Bwedge xnotin Cright)right]veeleft[xin Bwedge xnotin Cwedge xnotin Aright] \
&iff left[xin Awedgeleft(xnotin Bvee xin Cright)right]veeleft[xnotin Awedge xin Bwedge xnotin Cright] \
&iff left[xin Awedge xnotin Bright]veeleft[xin Awedge xin Cright]veeleft[xnotin Awedge xin Bwedge xnotin Cright].
end{split}$$
Each part of the last expression describes a piece in the Venn diagram for $Abigtriangleup(Bsetminus C)$. In particular, $left[xin Awedge xnotin Bright]$ is $Asetminus B$, but it includes two pieces of the Venn diagram, and we can see that we'd like to separate them:
$$begin{split}
xin Awedge xnotin B &iff xin Awedge xnotin Bwedgeleft(xin Cvee xnotin Cright) \
&iff left[xin Awedge xnotin Bwedge xin Cright]veeleft[xin Awedge xnotin Bwedge xnotin Cright].
end{split}$$
If we put it back into the last step above, then we'll get that $xin Abigtriangleup(Bsetminus C)$ iff
$$left[xin Awedge xnotin Bwedge xin Cright]veeleft[xin Awedge xnotin Bwedge xnotin Cright]veeleft[xin Awedge xin Cright]veeleft[xnotin Awedge xin Bwedge xnotin Cright].$$
Moreover, we should observe that $left[xin Awedge xnotin Bwedge xin Cright]$ is "included" in, i.e. describes a subset of, $left[xin Awedge xin Cright]$. We can work it out using or and and in the same fashion (I'll leave that to you), but the bottom line is that the former ends up being "consumed" by the latter, so we'll get that $xin Abigtriangleup(Bsetminus C)$ iff
$$color{blue}{left[xin Awedge xnotin Bwedge xnotin Cright]veeleft[xnotin Awedge xin Bwedge xnotin Cright]}veecolor{red}{left[xin Awedge xin Cright]}.$$
Note that the blue parts are precisely what is included on the other diagram, while the red part describes $Acap C$ — which is precisely what distinguishes the two diagrams.
Now, if we work out $(Abigtriangleup B)setminus C$ in a similar fashion, we'll end precisely with the blue expression. That will show that the two are equal iff $Acap C=varnothing$.
NOTE: I do realize that this solution seems lengthy and difficult. Yes, it is lengthy and tedious, but it's completely dictated by the pictures. From the visual point of view, I'd say it's pretty straightforward.
answered Jan 5 at 18:13
zipirovichzipirovich
11.2k11631
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$begingroup$
Thank you so much! Now I understand it. I tried a similar way but in the end I didn't know that I can use $land (x in C lor x notin C)$.
$endgroup$
– Nemanja Djordjevic
Jan 8 at 1:19
add a comment |
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We wish to prove the following theorem:
$$
Atriangle(Bsetminus C) = (Atriangle B)setminus C quad Leftrightarrow quad Acap C=emptyset
$$
I don't particularly like obfuscating what is going on behind the symmetric difference and set-minus operators, so let us unwind that a bit:
$$
[Asetminus(Bsetminus C)]cup[(Bsetminus C)setminus A] = ([Asetminus B] cup [Bsetminus A])setminus C quad Leftrightarrow quad Acap C=emptyset
$$
$$
[Acap(Bcap C^c)^c]cup[(Bcap C^c)cap A^c] = ([Acap B^c] cup [Bcap A^c])cap C^c quad Leftrightarrow quad Acap C=emptyset
$$
$$
[Acap(B^ccup C)]cup[(Bcap C^c)cap A^c] = ([Acap B^c] cup [Bcap A^c])cap C^c quad Leftrightarrow quad Acap C=emptyset
$$
Now, if you squint really hard at that last line, you will see that the $Leftarrow$ direction is incredibly easy to prove! Just distribute various intersections over various unions and utilize the hypothesis that $Acap C=emptyset$ (which is equivalent to $Acap C^c=A$). You will be left with $X=X$, which is true.
For the $Rightarrow$ direction, things are a bit harder. I would recommend approaching this via the contrapositive; assume that $Acap Cneqemptyset$ and try and prove that $Atriangle(Bsetminus C) neq (Atriangle B)setminus C$. If you do this, then again we find that the proof is very straightforaward:
$$
[Acap(B^ccup C)]cup[(Bcap C^c)cap A^c] stackrel{?}{=} ([Acap B^c] cup [Bcap A^c])cap C^c
$$
$$
[Acap B^c]cup[Acap C]cup[Bcap A^ccap C^c] stackrel{?}{=} [Acap B^ccap C^c] cup [Bcap A^ccap C^c]
$$
On the RHS we have the union of two disjoint sets. One of those sets also appears on the LHS, but the other is a proper subset of the first two bracketed sets on the LHS. Thus, equality is impossible!
This answer is not completely rigorous, because I got lazy and didn't want to name the element that belongs to the LHS and not the RHS for the inequality argument. However, it should be a good enough starting point for you to finish this completely.
answered Jan 5 at 2:47
ItsJustTranscendenceBroItsJustTranscendenceBro
1712
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this supposed to be an easy proof :(
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– Nemanja Djordjevic
Jan 5 at 3:00
1
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@NemanjaDjordjevic To be fair, this is an easy argument. It just happens to also look long and messy. Perhaps there is a slick and fast way, but for me this makes the most sense and is the easiest proof I could come up with.
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– ItsJustTranscendenceBro
Jan 5 at 3:01
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Thank you very much!
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– Nemanja Djordjevic
Jan 5 at 3:03
add a comment |
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No that step is correct because in the case which $$[x in (A backslash B) backslash C)] vee [xin(B backslash A) backslash C]$$ $x$ doesn't belong to $C$ in either cases so that you can kick $C$ out of both the bracket cases. Note that this only proves that $$Atriangle(Bbackslash C)subseteq(Atriangle B)backslash C $$you will also need a converse proof on this theorem (which is as the same proof of direct side of the theorem in reverse direction).
Alternative proof
$$Atriangle (Bbackslash C){=Big[(B-C)-ABig]cupBig[A-(B-C)Big]\=Big[Bcap C'cap A']cupBig[Acap(Bcap C')'Big]\=Big[Bcap C'cap A']cupBig[Acap(B'cup C)Big]}$$and $$(Atriangle B)backslash C{=Big[(A-B)cup(B-A)Big]cap C'\=Big[Acap B'cap C'Big]cup Big[Bcap C'cap A'Big]\=Big[Acap (Bcup C)'Big]cup Big[Bcap C'cap A'Big]}$$then we need to show that $$Acap (Bcup C)'=Acap(B'cup C)iff Acap C=emptyset$$from the other side $$Big[Acap (Bcup C)'Big]cupBig[Acap CBig]{=AcapBig[(Bcup C)'cup CBig]\=AcapBig[(B'cap C')cup CBig]\=AcapBig[(B'cup C)cap (C'cup C)Big]\=AcapBig[(B'cup C)cap mathrm{U}Big]\=Acap(B'cup C)}$$where $mathrm{U}$ is the Universal Set. Also $$Big[Acap (Bcup C)'Big]capBig[Acap CBig]{=Big[Acap B'cap C'Big]capBig[Acap CBig]\=Acap B'cap C'cap C\=emptyset}$$therefore $$Acap (Bcup C)'=Acap(B'cup C)$$if and only if $$ Acap C=emptyset$$and the proof is complete.
answered Jan 5 at 4:03
Mostafa AyazMostafa Ayaz
15.4k3939
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What tells us that $xin(B backslash A) backslash C$ this is true?
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– Nemanja Djordjevic
Jan 5 at 4:12
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It means that $x$ belongs to $B$ but not to either $A$ or $C$. Another way of expressing it is as $$xin Bcap A'cap C'$$
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– Mostafa Ayaz
Jan 5 at 4:16
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But x belongs to B
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– Nemanja Djordjevic
Jan 5 at 4:17
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Sorry for the typo.........
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– Mostafa Ayaz
Jan 5 at 4:18
1
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Yes because the taking complementary then intersection operation is associative i.e. $$((A-B_1)-B_2)-cdots - B_n=Abigcap_{i=1}^{n}B_i'=Acapleft(bigcup_{i=1}^{n}B_iright)'$$
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– Mostafa Ayaz
Jan 5 at 4:29
|
show 3 more comments
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OK, we want to show
$$A Delta (B setminus C) = (A Delta B)setminus C Leftrightarrow A cap C = emptyset$$
Assume we have $A cap C neq emptyset$ (so trying for the contrapositive, which seems easier as it gives us something concrete to work with). Hence we have some $p in A cap C$. Case 1: $p in B$, then $p notin (B setminus C)$ (as $p in C$) but also $p in A$, so $p in A Delta (B setminus C)$ but on the other hand $p notin A Delta B$ so $p notin (A Delta B) setminus C$ so we have disproved the LHS identity in this case. Case 2: $p notin B$, then $p in A Delta B$ (and still $p in C$) so $p notin (A Delta B) setminus C$, and $p notin B setminus C$ and still $p in A$ so $p in A Delta (Bsetminus C)$ as in case 1, we have disproved the identity again.
Now assume that $A cap C = emptyset$ holds. We want to show the LHS identity which is really two inclusions. So suppose $x in A Delta (B setminus C)$, so case 1: $x in A$ and $x notin B setminus C$. As $x notin C$ (as otherwise $A cap C neq emptyset$) we know also that $x notin B$ (or else $x in B setminus C)$ after all) and so $x in (A Delta B)setminus C$, so the left to right inclusion holds for case 1. Case 2: $x notin A$ and $x in B setminus C$, and then $x in B$ and $x notin C$ and so $x in (A Delta B)setminus C$ and so the left to right inclusion also holds in case 2. So we're done with that inclusion. Now suppose that $x in (A Delta B) setminus C$ so $x in A Delta B$ and $x notin C$. Again 2 cases: case 1: $x in A$, $x notin B$, but then $x notin B setminus C$ and so $x in A Delta (B setminus C)$ and the right to left inclusion holds. Case 2: $x notin A$ and $x in B$, and then $x in B setminus C$ and so $x in A Delta (B setminus C)$ and the right to left inclusion holds (without even using $A cap C = emptyset$, so that inclusion always holds). So we have the LHS equality when $A$ and $C$ are disjoint.
answered Jan 5 at 6:43
Henno BrandsmaHenno Brandsma
106k347114
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4 Answers
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4 Answers
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$begingroup$
These are the Venn diagrams for both expressions. As we can see, they differ precisely by $Acap C$, which is why $Acap C=varnothing$ is required to make them equal. Also, notice that $(Abigtriangleup B)setminus Csubseteq Abigtriangleup(Bsetminus C)$ always; it's only the opposite inclusion that needs $Acap C=varnothing$.
If you're expected to prove the claim using just or and and logical connectors, then your first steps look good, but then I guess you should continue expanding all the definitions. Let's see how that looks for the inclusion $Abigtriangleup(Bsetminus C)subseteq(Abigtriangleup B)setminus C)$ that you started working on.
$$begin{split}
xin Abigtriangleup(Bsetminus C) &iff left[xin Asetminus(Bsetminus C)right]veeleft[xin(Bsetminus C)setminus Aright] \
&iff left[xin Awedge xnotin(Bsetminus C)right]veeleft[xin(Bsetminus C)wedge xnotin Aright] \
&iff left[xin Awedgenegleft(xin Bwedge xnotin Cright)right]veeleft[xin Bwedge xnotin Cwedge xnotin Aright] \
&iff left[xin Awedgeleft(xnotin Bvee xin Cright)right]veeleft[xnotin Awedge xin Bwedge xnotin Cright] \
&iff left[xin Awedge xnotin Bright]veeleft[xin Awedge xin Cright]veeleft[xnotin Awedge xin Bwedge xnotin Cright].
end{split}$$
Each part of the last expression describes a piece in the Venn diagram for $Abigtriangleup(Bsetminus C)$. In particular, $left[xin Awedge xnotin Bright]$ is $Asetminus B$, but it includes two pieces of the Venn diagram, and we can see that we'd like to separate them:
$$begin{split}
xin Awedge xnotin B &iff xin Awedge xnotin Bwedgeleft(xin Cvee xnotin Cright) \
&iff left[xin Awedge xnotin Bwedge xin Cright]veeleft[xin Awedge xnotin Bwedge xnotin Cright].
end{split}$$
If we put it back into the last step above, then we'll get that $xin Abigtriangleup(Bsetminus C)$ iff
$$left[xin Awedge xnotin Bwedge xin Cright]veeleft[xin Awedge xnotin Bwedge xnotin Cright]veeleft[xin Awedge xin Cright]veeleft[xnotin Awedge xin Bwedge xnotin Cright].$$
Moreover, we should observe that $left[xin Awedge xnotin Bwedge xin Cright]$ is "included" in, i.e. describes a subset of, $left[xin Awedge xin Cright]$. We can work it out using or and and in the same fashion (I'll leave that to you), but the bottom line is that the former ends up being "consumed" by the latter, so we'll get that $xin Abigtriangleup(Bsetminus C)$ iff
$$color{blue}{left[xin Awedge xnotin Bwedge xnotin Cright]veeleft[xnotin Awedge xin Bwedge xnotin Cright]}veecolor{red}{left[xin Awedge xin Cright]}.$$
Note that the blue parts are precisely what is included on the other diagram, while the red part describes $Acap C$ — which is precisely what distinguishes the two diagrams.
Now, if we work out $(Abigtriangleup B)setminus C$ in a similar fashion, we'll end precisely with the blue expression. That will show that the two are equal iff $Acap C=varnothing$.
NOTE: I do realize that this solution seems lengthy and difficult. Yes, it is lengthy and tedious, but it's completely dictated by the pictures. From the visual point of view, I'd say it's pretty straightforward.
answered Jan 5 at 18:13
zipirovichzipirovich
11.2k11631
$endgroup$
$begingroup$
Thank you so much! Now I understand it. I tried a similar way but in the end I didn't know that I can use $land (x in C lor x notin C)$.
$endgroup$
– Nemanja Djordjevic
Jan 8 at 1:19
add a comment |
$begingroup$
These are the Venn diagrams for both expressions. As we can see, they differ precisely by $Acap C$, which is why $Acap C=varnothing$ is required to make them equal. Also, notice that $(Abigtriangleup B)setminus Csubseteq Abigtriangleup(Bsetminus C)$ always; it's only the opposite inclusion that needs $Acap C=varnothing$.
If you're expected to prove the claim using just or and and logical connectors, then your first steps look good, but then I guess you should continue expanding all the definitions. Let's see how that looks for the inclusion $Abigtriangleup(Bsetminus C)subseteq(Abigtriangleup B)setminus C)$ that you started working on.
$$begin{split}
xin Abigtriangleup(Bsetminus C) &iff left[xin Asetminus(Bsetminus C)right]veeleft[xin(Bsetminus C)setminus Aright] \
&iff left[xin Awedge xnotin(Bsetminus C)right]veeleft[xin(Bsetminus C)wedge xnotin Aright] \
&iff left[xin Awedgenegleft(xin Bwedge xnotin Cright)right]veeleft[xin Bwedge xnotin Cwedge xnotin Aright] \
&iff left[xin Awedgeleft(xnotin Bvee xin Cright)right]veeleft[xnotin Awedge xin Bwedge xnotin Cright] \
&iff left[xin Awedge xnotin Bright]veeleft[xin Awedge xin Cright]veeleft[xnotin Awedge xin Bwedge xnotin Cright].
end{split}$$
Each part of the last expression describes a piece in the Venn diagram for $Abigtriangleup(Bsetminus C)$. In particular, $left[xin Awedge xnotin Bright]$ is $Asetminus B$, but it includes two pieces of the Venn diagram, and we can see that we'd like to separate them:
$$begin{split}
xin Awedge xnotin B &iff xin Awedge xnotin Bwedgeleft(xin Cvee xnotin Cright) \
&iff left[xin Awedge xnotin Bwedge xin Cright]veeleft[xin Awedge xnotin Bwedge xnotin Cright].
end{split}$$
If we put it back into the last step above, then we'll get that $xin Abigtriangleup(Bsetminus C)$ iff
$$left[xin Awedge xnotin Bwedge xin Cright]veeleft[xin Awedge xnotin Bwedge xnotin Cright]veeleft[xin Awedge xin Cright]veeleft[xnotin Awedge xin Bwedge xnotin Cright].$$
Moreover, we should observe that $left[xin Awedge xnotin Bwedge xin Cright]$ is "included" in, i.e. describes a subset of, $left[xin Awedge xin Cright]$. We can work it out using or and and in the same fashion (I'll leave that to you), but the bottom line is that the former ends up being "consumed" by the latter, so we'll get that $xin Abigtriangleup(Bsetminus C)$ iff
$$color{blue}{left[xin Awedge xnotin Bwedge xnotin Cright]veeleft[xnotin Awedge xin Bwedge xnotin Cright]}veecolor{red}{left[xin Awedge xin Cright]}.$$
Note that the blue parts are precisely what is included on the other diagram, while the red part describes $Acap C$ — which is precisely what distinguishes the two diagrams.
Now, if we work out $(Abigtriangleup B)setminus C$ in a similar fashion, we'll end precisely with the blue expression. That will show that the two are equal iff $Acap C=varnothing$.
NOTE: I do realize that this solution seems lengthy and difficult. Yes, it is lengthy and tedious, but it's completely dictated by the pictures. From the visual point of view, I'd say it's pretty straightforward.
answered Jan 5 at 18:13
zipirovichzipirovich
11.2k11631
$endgroup$
$begingroup$
Thank you so much! Now I understand it. I tried a similar way but in the end I didn't know that I can use $land (x in C lor x notin C)$.
$endgroup$
– Nemanja Djordjevic
Jan 8 at 1:19
add a comment |
$begingroup$
These are the Venn diagrams for both expressions. As we can see, they differ precisely by $Acap C$, which is why $Acap C=varnothing$ is required to make them equal. Also, notice that $(Abigtriangleup B)setminus Csubseteq Abigtriangleup(Bsetminus C)$ always; it's only the opposite inclusion that needs $Acap C=varnothing$.
If you're expected to prove the claim using just or and and logical connectors, then your first steps look good, but then I guess you should continue expanding all the definitions. Let's see how that looks for the inclusion $Abigtriangleup(Bsetminus C)subseteq(Abigtriangleup B)setminus C)$ that you started working on.
$$begin{split}
xin Abigtriangleup(Bsetminus C) &iff left[xin Asetminus(Bsetminus C)right]veeleft[xin(Bsetminus C)setminus Aright] \
&iff left[xin Awedge xnotin(Bsetminus C)right]veeleft[xin(Bsetminus C)wedge xnotin Aright] \
&iff left[xin Awedgenegleft(xin Bwedge xnotin Cright)right]veeleft[xin Bwedge xnotin Cwedge xnotin Aright] \
&iff left[xin Awedgeleft(xnotin Bvee xin Cright)right]veeleft[xnotin Awedge xin Bwedge xnotin Cright] \
&iff left[xin Awedge xnotin Bright]veeleft[xin Awedge xin Cright]veeleft[xnotin Awedge xin Bwedge xnotin Cright].
end{split}$$
Each part of the last expression describes a piece in the Venn diagram for $Abigtriangleup(Bsetminus C)$. In particular, $left[xin Awedge xnotin Bright]$ is $Asetminus B$, but it includes two pieces of the Venn diagram, and we can see that we'd like to separate them:
$$begin{split}
xin Awedge xnotin B &iff xin Awedge xnotin Bwedgeleft(xin Cvee xnotin Cright) \
&iff left[xin Awedge xnotin Bwedge xin Cright]veeleft[xin Awedge xnotin Bwedge xnotin Cright].
end{split}$$
If we put it back into the last step above, then we'll get that $xin Abigtriangleup(Bsetminus C)$ iff
$$left[xin Awedge xnotin Bwedge xin Cright]veeleft[xin Awedge xnotin Bwedge xnotin Cright]veeleft[xin Awedge xin Cright]veeleft[xnotin Awedge xin Bwedge xnotin Cright].$$
Moreover, we should observe that $left[xin Awedge xnotin Bwedge xin Cright]$ is "included" in, i.e. describes a subset of, $left[xin Awedge xin Cright]$. We can work it out using or and and in the same fashion (I'll leave that to you), but the bottom line is that the former ends up being "consumed" by the latter, so we'll get that $xin Abigtriangleup(Bsetminus C)$ iff
$$color{blue}{left[xin Awedge xnotin Bwedge xnotin Cright]veeleft[xnotin Awedge xin Bwedge xnotin Cright]}veecolor{red}{left[xin Awedge xin Cright]}.$$
Note that the blue parts are precisely what is included on the other diagram, while the red part describes $Acap C$ — which is precisely what distinguishes the two diagrams.
Now, if we work out $(Abigtriangleup B)setminus C$ in a similar fashion, we'll end precisely with the blue expression. That will show that the two are equal iff $Acap C=varnothing$.
NOTE: I do realize that this solution seems lengthy and difficult. Yes, it is lengthy and tedious, but it's completely dictated by the pictures. From the visual point of view, I'd say it's pretty straightforward.
answered Jan 5 at 18:13
zipirovichzipirovich
11.2k11631
$endgroup$
These are the Venn diagrams for both expressions. As we can see, they differ precisely by $Acap C$, which is why $Acap C=varnothing$ is required to make them equal. Also, notice that $(Abigtriangleup B)setminus Csubseteq Abigtriangleup(Bsetminus C)$ always; it's only the opposite inclusion that needs $Acap C=varnothing$.
If you're expected to prove the claim using just or and and logical connectors, then your first steps look good, but then I guess you should continue expanding all the definitions. Let's see how that looks for the inclusion $Abigtriangleup(Bsetminus C)subseteq(Abigtriangleup B)setminus C)$ that you started working on.
$$begin{split}
xin Abigtriangleup(Bsetminus C) &iff left[xin Asetminus(Bsetminus C)right]veeleft[xin(Bsetminus C)setminus Aright] \
&iff left[xin Awedge xnotin(Bsetminus C)right]veeleft[xin(Bsetminus C)wedge xnotin Aright] \
&iff left[xin Awedgenegleft(xin Bwedge xnotin Cright)right]veeleft[xin Bwedge xnotin Cwedge xnotin Aright] \
&iff left[xin Awedgeleft(xnotin Bvee xin Cright)right]veeleft[xnotin Awedge xin Bwedge xnotin Cright] \
&iff left[xin Awedge xnotin Bright]veeleft[xin Awedge xin Cright]veeleft[xnotin Awedge xin Bwedge xnotin Cright].
end{split}$$
Each part of the last expression describes a piece in the Venn diagram for $Abigtriangleup(Bsetminus C)$. In particular, $left[xin Awedge xnotin Bright]$ is $Asetminus B$, but it includes two pieces of the Venn diagram, and we can see that we'd like to separate them:
$$begin{split}
xin Awedge xnotin B &iff xin Awedge xnotin Bwedgeleft(xin Cvee xnotin Cright) \
&iff left[xin Awedge xnotin Bwedge xin Cright]veeleft[xin Awedge xnotin Bwedge xnotin Cright].
end{split}$$
If we put it back into the last step above, then we'll get that $xin Abigtriangleup(Bsetminus C)$ iff
$$left[xin Awedge xnotin Bwedge xin Cright]veeleft[xin Awedge xnotin Bwedge xnotin Cright]veeleft[xin Awedge xin Cright]veeleft[xnotin Awedge xin Bwedge xnotin Cright].$$
Moreover, we should observe that $left[xin Awedge xnotin Bwedge xin Cright]$ is "included" in, i.e. describes a subset of, $left[xin Awedge xin Cright]$. We can work it out using or and and in the same fashion (I'll leave that to you), but the bottom line is that the former ends up being "consumed" by the latter, so we'll get that $xin Abigtriangleup(Bsetminus C)$ iff
$$color{blue}{left[xin Awedge xnotin Bwedge xnotin Cright]veeleft[xnotin Awedge xin Bwedge xnotin Cright]}veecolor{red}{left[xin Awedge xin Cright]}.$$
Note that the blue parts are precisely what is included on the other diagram, while the red part describes $Acap C$ — which is precisely what distinguishes the two diagrams.
Now, if we work out $(Abigtriangleup B)setminus C$ in a similar fashion, we'll end precisely with the blue expression. That will show that the two are equal iff $Acap C=varnothing$.
NOTE: I do realize that this solution seems lengthy and difficult. Yes, it is lengthy and tedious, but it's completely dictated by the pictures. From the visual point of view, I'd say it's pretty straightforward.
answered Jan 5 at 18:13
zipirovichzipirovich
11.2k11631
edited Jan 8 at 1:45
answered Jan 5 at 18:13
zipirovichzipirovich
11.2k11631
answered Jan 5 at 18:13
zipirovichzipirovich
11.2k11631
answered Jan 5 at 18:13
zipirovichzipirovich
11.2k11631
11.2k11631
$begingroup$
Thank you so much! Now I understand it. I tried a similar way but in the end I didn't know that I can use $land (x in C lor x notin C)$.
$endgroup$
– Nemanja Djordjevic
Jan 8 at 1:19
add a comment |
$begingroup$
Thank you so much! Now I understand it. I tried a similar way but in the end I didn't know that I can use $land (x in C lor x notin C)$.
$endgroup$
– Nemanja Djordjevic
Jan 8 at 1:19
$begingroup$
Thank you so much! Now I understand it. I tried a similar way but in the end I didn't know that I can use $land (x in C lor x notin C)$.
$endgroup$
– Nemanja Djordjevic
Jan 8 at 1:19
$begingroup$
Thank you so much! Now I understand it. I tried a similar way but in the end I didn't know that I can use $land (x in C lor x notin C)$.
$endgroup$
– Nemanja Djordjevic
Jan 8 at 1:19
add a comment |
$begingroup$
We wish to prove the following theorem:
$$
Atriangle(Bsetminus C) = (Atriangle B)setminus C quad Leftrightarrow quad Acap C=emptyset
$$
I don't particularly like obfuscating what is going on behind the symmetric difference and set-minus operators, so let us unwind that a bit:
$$
[Asetminus(Bsetminus C)]cup[(Bsetminus C)setminus A] = ([Asetminus B] cup [Bsetminus A])setminus C quad Leftrightarrow quad Acap C=emptyset
$$
$$
[Acap(Bcap C^c)^c]cup[(Bcap C^c)cap A^c] = ([Acap B^c] cup [Bcap A^c])cap C^c quad Leftrightarrow quad Acap C=emptyset
$$
$$
[Acap(B^ccup C)]cup[(Bcap C^c)cap A^c] = ([Acap B^c] cup [Bcap A^c])cap C^c quad Leftrightarrow quad Acap C=emptyset
$$
Now, if you squint really hard at that last line, you will see that the $Leftarrow$ direction is incredibly easy to prove! Just distribute various intersections over various unions and utilize the hypothesis that $Acap C=emptyset$ (which is equivalent to $Acap C^c=A$). You will be left with $X=X$, which is true.
For the $Rightarrow$ direction, things are a bit harder. I would recommend approaching this via the contrapositive; assume that $Acap Cneqemptyset$ and try and prove that $Atriangle(Bsetminus C) neq (Atriangle B)setminus C$. If you do this, then again we find that the proof is very straightforaward:
$$
[Acap(B^ccup C)]cup[(Bcap C^c)cap A^c] stackrel{?}{=} ([Acap B^c] cup [Bcap A^c])cap C^c
$$
$$
[Acap B^c]cup[Acap C]cup[Bcap A^ccap C^c] stackrel{?}{=} [Acap B^ccap C^c] cup [Bcap A^ccap C^c]
$$
On the RHS we have the union of two disjoint sets. One of those sets also appears on the LHS, but the other is a proper subset of the first two bracketed sets on the LHS. Thus, equality is impossible!
This answer is not completely rigorous, because I got lazy and didn't want to name the element that belongs to the LHS and not the RHS for the inequality argument. However, it should be a good enough starting point for you to finish this completely.
answered Jan 5 at 2:47
ItsJustTranscendenceBroItsJustTranscendenceBro
1712
$endgroup$
$begingroup$
this supposed to be an easy proof :(
$endgroup$
– Nemanja Djordjevic
Jan 5 at 3:00
1
$begingroup$
@NemanjaDjordjevic To be fair, this is an easy argument. It just happens to also look long and messy. Perhaps there is a slick and fast way, but for me this makes the most sense and is the easiest proof I could come up with.
$endgroup$
– ItsJustTranscendenceBro
Jan 5 at 3:01
$begingroup$
Thank you very much!
$endgroup$
– Nemanja Djordjevic
Jan 5 at 3:03
add a comment |
$begingroup$
We wish to prove the following theorem:
$$
Atriangle(Bsetminus C) = (Atriangle B)setminus C quad Leftrightarrow quad Acap C=emptyset
$$
I don't particularly like obfuscating what is going on behind the symmetric difference and set-minus operators, so let us unwind that a bit:
$$
[Asetminus(Bsetminus C)]cup[(Bsetminus C)setminus A] = ([Asetminus B] cup [Bsetminus A])setminus C quad Leftrightarrow quad Acap C=emptyset
$$
$$
[Acap(Bcap C^c)^c]cup[(Bcap C^c)cap A^c] = ([Acap B^c] cup [Bcap A^c])cap C^c quad Leftrightarrow quad Acap C=emptyset
$$
$$
[Acap(B^ccup C)]cup[(Bcap C^c)cap A^c] = ([Acap B^c] cup [Bcap A^c])cap C^c quad Leftrightarrow quad Acap C=emptyset
$$
Now, if you squint really hard at that last line, you will see that the $Leftarrow$ direction is incredibly easy to prove! Just distribute various intersections over various unions and utilize the hypothesis that $Acap C=emptyset$ (which is equivalent to $Acap C^c=A$). You will be left with $X=X$, which is true.
For the $Rightarrow$ direction, things are a bit harder. I would recommend approaching this via the contrapositive; assume that $Acap Cneqemptyset$ and try and prove that $Atriangle(Bsetminus C) neq (Atriangle B)setminus C$. If you do this, then again we find that the proof is very straightforaward:
$$
[Acap(B^ccup C)]cup[(Bcap C^c)cap A^c] stackrel{?}{=} ([Acap B^c] cup [Bcap A^c])cap C^c
$$
$$
[Acap B^c]cup[Acap C]cup[Bcap A^ccap C^c] stackrel{?}{=} [Acap B^ccap C^c] cup [Bcap A^ccap C^c]
$$
On the RHS we have the union of two disjoint sets. One of those sets also appears on the LHS, but the other is a proper subset of the first two bracketed sets on the LHS. Thus, equality is impossible!
This answer is not completely rigorous, because I got lazy and didn't want to name the element that belongs to the LHS and not the RHS for the inequality argument. However, it should be a good enough starting point for you to finish this completely.
answered Jan 5 at 2:47
ItsJustTranscendenceBroItsJustTranscendenceBro
1712
$endgroup$
$begingroup$
this supposed to be an easy proof :(
$endgroup$
– Nemanja Djordjevic
Jan 5 at 3:00
1
$begingroup$
@NemanjaDjordjevic To be fair, this is an easy argument. It just happens to also look long and messy. Perhaps there is a slick and fast way, but for me this makes the most sense and is the easiest proof I could come up with.
$endgroup$
– ItsJustTranscendenceBro
Jan 5 at 3:01
$begingroup$
Thank you very much!
$endgroup$
– Nemanja Djordjevic
Jan 5 at 3:03
add a comment |
$begingroup$
We wish to prove the following theorem:
$$
Atriangle(Bsetminus C) = (Atriangle B)setminus C quad Leftrightarrow quad Acap C=emptyset
$$
I don't particularly like obfuscating what is going on behind the symmetric difference and set-minus operators, so let us unwind that a bit:
$$
[Asetminus(Bsetminus C)]cup[(Bsetminus C)setminus A] = ([Asetminus B] cup [Bsetminus A])setminus C quad Leftrightarrow quad Acap C=emptyset
$$
$$
[Acap(Bcap C^c)^c]cup[(Bcap C^c)cap A^c] = ([Acap B^c] cup [Bcap A^c])cap C^c quad Leftrightarrow quad Acap C=emptyset
$$
$$
[Acap(B^ccup C)]cup[(Bcap C^c)cap A^c] = ([Acap B^c] cup [Bcap A^c])cap C^c quad Leftrightarrow quad Acap C=emptyset
$$
Now, if you squint really hard at that last line, you will see that the $Leftarrow$ direction is incredibly easy to prove! Just distribute various intersections over various unions and utilize the hypothesis that $Acap C=emptyset$ (which is equivalent to $Acap C^c=A$). You will be left with $X=X$, which is true.
For the $Rightarrow$ direction, things are a bit harder. I would recommend approaching this via the contrapositive; assume that $Acap Cneqemptyset$ and try and prove that $Atriangle(Bsetminus C) neq (Atriangle B)setminus C$. If you do this, then again we find that the proof is very straightforaward:
$$
[Acap(B^ccup C)]cup[(Bcap C^c)cap A^c] stackrel{?}{=} ([Acap B^c] cup [Bcap A^c])cap C^c
$$
$$
[Acap B^c]cup[Acap C]cup[Bcap A^ccap C^c] stackrel{?}{=} [Acap B^ccap C^c] cup [Bcap A^ccap C^c]
$$
On the RHS we have the union of two disjoint sets. One of those sets also appears on the LHS, but the other is a proper subset of the first two bracketed sets on the LHS. Thus, equality is impossible!
This answer is not completely rigorous, because I got lazy and didn't want to name the element that belongs to the LHS and not the RHS for the inequality argument. However, it should be a good enough starting point for you to finish this completely.
answered Jan 5 at 2:47
ItsJustTranscendenceBroItsJustTranscendenceBro
1712
$endgroup$
We wish to prove the following theorem:
$$
Atriangle(Bsetminus C) = (Atriangle B)setminus C quad Leftrightarrow quad Acap C=emptyset
$$
I don't particularly like obfuscating what is going on behind the symmetric difference and set-minus operators, so let us unwind that a bit:
$$
[Asetminus(Bsetminus C)]cup[(Bsetminus C)setminus A] = ([Asetminus B] cup [Bsetminus A])setminus C quad Leftrightarrow quad Acap C=emptyset
$$
$$
[Acap(Bcap C^c)^c]cup[(Bcap C^c)cap A^c] = ([Acap B^c] cup [Bcap A^c])cap C^c quad Leftrightarrow quad Acap C=emptyset
$$
$$
[Acap(B^ccup C)]cup[(Bcap C^c)cap A^c] = ([Acap B^c] cup [Bcap A^c])cap C^c quad Leftrightarrow quad Acap C=emptyset
$$
Now, if you squint really hard at that last line, you will see that the $Leftarrow$ direction is incredibly easy to prove! Just distribute various intersections over various unions and utilize the hypothesis that $Acap C=emptyset$ (which is equivalent to $Acap C^c=A$). You will be left with $X=X$, which is true.
For the $Rightarrow$ direction, things are a bit harder. I would recommend approaching this via the contrapositive; assume that $Acap Cneqemptyset$ and try and prove that $Atriangle(Bsetminus C) neq (Atriangle B)setminus C$. If you do this, then again we find that the proof is very straightforaward:
$$
[Acap(B^ccup C)]cup[(Bcap C^c)cap A^c] stackrel{?}{=} ([Acap B^c] cup [Bcap A^c])cap C^c
$$
$$
[Acap B^c]cup[Acap C]cup[Bcap A^ccap C^c] stackrel{?}{=} [Acap B^ccap C^c] cup [Bcap A^ccap C^c]
$$
On the RHS we have the union of two disjoint sets. One of those sets also appears on the LHS, but the other is a proper subset of the first two bracketed sets on the LHS. Thus, equality is impossible!
This answer is not completely rigorous, because I got lazy and didn't want to name the element that belongs to the LHS and not the RHS for the inequality argument. However, it should be a good enough starting point for you to finish this completely.
answered Jan 5 at 2:47
ItsJustTranscendenceBroItsJustTranscendenceBro
1712
answered Jan 5 at 2:47
ItsJustTranscendenceBroItsJustTranscendenceBro
1712
answered Jan 5 at 2:47
ItsJustTranscendenceBroItsJustTranscendenceBro
1712
answered Jan 5 at 2:47
ItsJustTranscendenceBroItsJustTranscendenceBro
1712
1712
$begingroup$
this supposed to be an easy proof :(
$endgroup$
– Nemanja Djordjevic
Jan 5 at 3:00
1
$begingroup$
@NemanjaDjordjevic To be fair, this is an easy argument. It just happens to also look long and messy. Perhaps there is a slick and fast way, but for me this makes the most sense and is the easiest proof I could come up with.
$endgroup$
– ItsJustTranscendenceBro
Jan 5 at 3:01
$begingroup$
Thank you very much!
$endgroup$
– Nemanja Djordjevic
Jan 5 at 3:03
add a comment |
$begingroup$
this supposed to be an easy proof :(
$endgroup$
– Nemanja Djordjevic
Jan 5 at 3:00
1
$begingroup$
@NemanjaDjordjevic To be fair, this is an easy argument. It just happens to also look long and messy. Perhaps there is a slick and fast way, but for me this makes the most sense and is the easiest proof I could come up with.
$endgroup$
– ItsJustTranscendenceBro
Jan 5 at 3:01
$begingroup$
Thank you very much!
$endgroup$
– Nemanja Djordjevic
Jan 5 at 3:03
$begingroup$
this supposed to be an easy proof :(
$endgroup$
– Nemanja Djordjevic
Jan 5 at 3:00
$begingroup$
this supposed to be an easy proof :(
$endgroup$
– Nemanja Djordjevic
Jan 5 at 3:00
1
1
$begingroup$
@NemanjaDjordjevic To be fair, this is an easy argument. It just happens to also look long and messy. Perhaps there is a slick and fast way, but for me this makes the most sense and is the easiest proof I could come up with.
$endgroup$
– ItsJustTranscendenceBro
Jan 5 at 3:01
$begingroup$
@NemanjaDjordjevic To be fair, this is an easy argument. It just happens to also look long and messy. Perhaps there is a slick and fast way, but for me this makes the most sense and is the easiest proof I could come up with.
$endgroup$
– ItsJustTranscendenceBro
Jan 5 at 3:01
$begingroup$
Thank you very much!
$endgroup$
– Nemanja Djordjevic
Jan 5 at 3:03
$begingroup$
Thank you very much!
$endgroup$
– Nemanja Djordjevic
Jan 5 at 3:03
add a comment |
$begingroup$
No that step is correct because in the case which $$[x in (A backslash B) backslash C)] vee [xin(B backslash A) backslash C]$$ $x$ doesn't belong to $C$ in either cases so that you can kick $C$ out of both the bracket cases. Note that this only proves that $$Atriangle(Bbackslash C)subseteq(Atriangle B)backslash C $$you will also need a converse proof on this theorem (which is as the same proof of direct side of the theorem in reverse direction).
Alternative proof
$$Atriangle (Bbackslash C){=Big[(B-C)-ABig]cupBig[A-(B-C)Big]\=Big[Bcap C'cap A']cupBig[Acap(Bcap C')'Big]\=Big[Bcap C'cap A']cupBig[Acap(B'cup C)Big]}$$and $$(Atriangle B)backslash C{=Big[(A-B)cup(B-A)Big]cap C'\=Big[Acap B'cap C'Big]cup Big[Bcap C'cap A'Big]\=Big[Acap (Bcup C)'Big]cup Big[Bcap C'cap A'Big]}$$then we need to show that $$Acap (Bcup C)'=Acap(B'cup C)iff Acap C=emptyset$$from the other side $$Big[Acap (Bcup C)'Big]cupBig[Acap CBig]{=AcapBig[(Bcup C)'cup CBig]\=AcapBig[(B'cap C')cup CBig]\=AcapBig[(B'cup C)cap (C'cup C)Big]\=AcapBig[(B'cup C)cap mathrm{U}Big]\=Acap(B'cup C)}$$where $mathrm{U}$ is the Universal Set. Also $$Big[Acap (Bcup C)'Big]capBig[Acap CBig]{=Big[Acap B'cap C'Big]capBig[Acap CBig]\=Acap B'cap C'cap C\=emptyset}$$therefore $$Acap (Bcup C)'=Acap(B'cup C)$$if and only if $$ Acap C=emptyset$$and the proof is complete.
answered Jan 5 at 4:03
Mostafa AyazMostafa Ayaz
15.4k3939
$endgroup$
$begingroup$
What tells us that $xin(B backslash A) backslash C$ this is true?
$endgroup$
– Nemanja Djordjevic
Jan 5 at 4:12
$begingroup$
It means that $x$ belongs to $B$ but not to either $A$ or $C$. Another way of expressing it is as $$xin Bcap A'cap C'$$
$endgroup$
– Mostafa Ayaz
Jan 5 at 4:16
$begingroup$
But x belongs to B
$endgroup$
– Nemanja Djordjevic
Jan 5 at 4:17
$begingroup$
Sorry for the typo.........
$endgroup$
– Mostafa Ayaz
Jan 5 at 4:18
1
$begingroup$
Yes because the taking complementary then intersection operation is associative i.e. $$((A-B_1)-B_2)-cdots - B_n=Abigcap_{i=1}^{n}B_i'=Acapleft(bigcup_{i=1}^{n}B_iright)'$$
$endgroup$
– Mostafa Ayaz
Jan 5 at 4:29
|
show 3 more comments
$begingroup$
No that step is correct because in the case which $$[x in (A backslash B) backslash C)] vee [xin(B backslash A) backslash C]$$ $x$ doesn't belong to $C$ in either cases so that you can kick $C$ out of both the bracket cases. Note that this only proves that $$Atriangle(Bbackslash C)subseteq(Atriangle B)backslash C $$you will also need a converse proof on this theorem (which is as the same proof of direct side of the theorem in reverse direction).
Alternative proof
$$Atriangle (Bbackslash C){=Big[(B-C)-ABig]cupBig[A-(B-C)Big]\=Big[Bcap C'cap A']cupBig[Acap(Bcap C')'Big]\=Big[Bcap C'cap A']cupBig[Acap(B'cup C)Big]}$$and $$(Atriangle B)backslash C{=Big[(A-B)cup(B-A)Big]cap C'\=Big[Acap B'cap C'Big]cup Big[Bcap C'cap A'Big]\=Big[Acap (Bcup C)'Big]cup Big[Bcap C'cap A'Big]}$$then we need to show that $$Acap (Bcup C)'=Acap(B'cup C)iff Acap C=emptyset$$from the other side $$Big[Acap (Bcup C)'Big]cupBig[Acap CBig]{=AcapBig[(Bcup C)'cup CBig]\=AcapBig[(B'cap C')cup CBig]\=AcapBig[(B'cup C)cap (C'cup C)Big]\=AcapBig[(B'cup C)cap mathrm{U}Big]\=Acap(B'cup C)}$$where $mathrm{U}$ is the Universal Set. Also $$Big[Acap (Bcup C)'Big]capBig[Acap CBig]{=Big[Acap B'cap C'Big]capBig[Acap CBig]\=Acap B'cap C'cap C\=emptyset}$$therefore $$Acap (Bcup C)'=Acap(B'cup C)$$if and only if $$ Acap C=emptyset$$and the proof is complete.
answered Jan 5 at 4:03
Mostafa AyazMostafa Ayaz
15.4k3939
$endgroup$
$begingroup$
What tells us that $xin(B backslash A) backslash C$ this is true?
$endgroup$
– Nemanja Djordjevic
Jan 5 at 4:12
$begingroup$
It means that $x$ belongs to $B$ but not to either $A$ or $C$. Another way of expressing it is as $$xin Bcap A'cap C'$$
$endgroup$
– Mostafa Ayaz
Jan 5 at 4:16
$begingroup$
But x belongs to B
$endgroup$
– Nemanja Djordjevic
Jan 5 at 4:17
$begingroup$
Sorry for the typo.........
$endgroup$
– Mostafa Ayaz
Jan 5 at 4:18
1
$begingroup$
Yes because the taking complementary then intersection operation is associative i.e. $$((A-B_1)-B_2)-cdots - B_n=Abigcap_{i=1}^{n}B_i'=Acapleft(bigcup_{i=1}^{n}B_iright)'$$
$endgroup$
– Mostafa Ayaz
Jan 5 at 4:29
|
show 3 more comments
$begingroup$
No that step is correct because in the case which $$[x in (A backslash B) backslash C)] vee [xin(B backslash A) backslash C]$$ $x$ doesn't belong to $C$ in either cases so that you can kick $C$ out of both the bracket cases. Note that this only proves that $$Atriangle(Bbackslash C)subseteq(Atriangle B)backslash C $$you will also need a converse proof on this theorem (which is as the same proof of direct side of the theorem in reverse direction).
Alternative proof
$$Atriangle (Bbackslash C){=Big[(B-C)-ABig]cupBig[A-(B-C)Big]\=Big[Bcap C'cap A']cupBig[Acap(Bcap C')'Big]\=Big[Bcap C'cap A']cupBig[Acap(B'cup C)Big]}$$and $$(Atriangle B)backslash C{=Big[(A-B)cup(B-A)Big]cap C'\=Big[Acap B'cap C'Big]cup Big[Bcap C'cap A'Big]\=Big[Acap (Bcup C)'Big]cup Big[Bcap C'cap A'Big]}$$then we need to show that $$Acap (Bcup C)'=Acap(B'cup C)iff Acap C=emptyset$$from the other side $$Big[Acap (Bcup C)'Big]cupBig[Acap CBig]{=AcapBig[(Bcup C)'cup CBig]\=AcapBig[(B'cap C')cup CBig]\=AcapBig[(B'cup C)cap (C'cup C)Big]\=AcapBig[(B'cup C)cap mathrm{U}Big]\=Acap(B'cup C)}$$where $mathrm{U}$ is the Universal Set. Also $$Big[Acap (Bcup C)'Big]capBig[Acap CBig]{=Big[Acap B'cap C'Big]capBig[Acap CBig]\=Acap B'cap C'cap C\=emptyset}$$therefore $$Acap (Bcup C)'=Acap(B'cup C)$$if and only if $$ Acap C=emptyset$$and the proof is complete.
answered Jan 5 at 4:03
Mostafa AyazMostafa Ayaz
15.4k3939
$endgroup$
No that step is correct because in the case which $$[x in (A backslash B) backslash C)] vee [xin(B backslash A) backslash C]$$ $x$ doesn't belong to $C$ in either cases so that you can kick $C$ out of both the bracket cases. Note that this only proves that $$Atriangle(Bbackslash C)subseteq(Atriangle B)backslash C $$you will also need a converse proof on this theorem (which is as the same proof of direct side of the theorem in reverse direction).
Alternative proof
$$Atriangle (Bbackslash C){=Big[(B-C)-ABig]cupBig[A-(B-C)Big]\=Big[Bcap C'cap A']cupBig[Acap(Bcap C')'Big]\=Big[Bcap C'cap A']cupBig[Acap(B'cup C)Big]}$$and $$(Atriangle B)backslash C{=Big[(A-B)cup(B-A)Big]cap C'\=Big[Acap B'cap C'Big]cup Big[Bcap C'cap A'Big]\=Big[Acap (Bcup C)'Big]cup Big[Bcap C'cap A'Big]}$$then we need to show that $$Acap (Bcup C)'=Acap(B'cup C)iff Acap C=emptyset$$from the other side $$Big[Acap (Bcup C)'Big]cupBig[Acap CBig]{=AcapBig[(Bcup C)'cup CBig]\=AcapBig[(B'cap C')cup CBig]\=AcapBig[(B'cup C)cap (C'cup C)Big]\=AcapBig[(B'cup C)cap mathrm{U}Big]\=Acap(B'cup C)}$$where $mathrm{U}$ is the Universal Set. Also $$Big[Acap (Bcup C)'Big]capBig[Acap CBig]{=Big[Acap B'cap C'Big]capBig[Acap CBig]\=Acap B'cap C'cap C\=emptyset}$$therefore $$Acap (Bcup C)'=Acap(B'cup C)$$if and only if $$ Acap C=emptyset$$and the proof is complete.
answered Jan 5 at 4:03
Mostafa AyazMostafa Ayaz
15.4k3939
edited Jan 5 at 5:44
answered Jan 5 at 4:03
Mostafa AyazMostafa Ayaz
15.4k3939
answered Jan 5 at 4:03
Mostafa AyazMostafa Ayaz
15.4k3939
answered Jan 5 at 4:03
Mostafa AyazMostafa Ayaz
15.4k3939
15.4k3939
$begingroup$
What tells us that $xin(B backslash A) backslash C$ this is true?
$endgroup$
– Nemanja Djordjevic
Jan 5 at 4:12
$begingroup$
It means that $x$ belongs to $B$ but not to either $A$ or $C$. Another way of expressing it is as $$xin Bcap A'cap C'$$
$endgroup$
– Mostafa Ayaz
Jan 5 at 4:16
$begingroup$
But x belongs to B
$endgroup$
– Nemanja Djordjevic
Jan 5 at 4:17
$begingroup$
Sorry for the typo.........
$endgroup$
– Mostafa Ayaz
Jan 5 at 4:18
1
$begingroup$
Yes because the taking complementary then intersection operation is associative i.e. $$((A-B_1)-B_2)-cdots - B_n=Abigcap_{i=1}^{n}B_i'=Acapleft(bigcup_{i=1}^{n}B_iright)'$$
$endgroup$
– Mostafa Ayaz
Jan 5 at 4:29
|
show 3 more comments
$begingroup$
What tells us that $xin(B backslash A) backslash C$ this is true?
$endgroup$
– Nemanja Djordjevic
Jan 5 at 4:12
$begingroup$
It means that $x$ belongs to $B$ but not to either $A$ or $C$. Another way of expressing it is as $$xin Bcap A'cap C'$$
$endgroup$
– Mostafa Ayaz
Jan 5 at 4:16
$begingroup$
But x belongs to B
$endgroup$
– Nemanja Djordjevic
Jan 5 at 4:17
$begingroup$
Sorry for the typo.........
$endgroup$
– Mostafa Ayaz
Jan 5 at 4:18
1
$begingroup$
Yes because the taking complementary then intersection operation is associative i.e. $$((A-B_1)-B_2)-cdots - B_n=Abigcap_{i=1}^{n}B_i'=Acapleft(bigcup_{i=1}^{n}B_iright)'$$
$endgroup$
– Mostafa Ayaz
Jan 5 at 4:29
$begingroup$
What tells us that $xin(B backslash A) backslash C$ this is true?
$endgroup$
– Nemanja Djordjevic
Jan 5 at 4:12
$begingroup$
What tells us that $xin(B backslash A) backslash C$ this is true?
$endgroup$
– Nemanja Djordjevic
Jan 5 at 4:12
$begingroup$
It means that $x$ belongs to $B$ but not to either $A$ or $C$. Another way of expressing it is as $$xin Bcap A'cap C'$$
$endgroup$
– Mostafa Ayaz
Jan 5 at 4:16
$begingroup$
It means that $x$ belongs to $B$ but not to either $A$ or $C$. Another way of expressing it is as $$xin Bcap A'cap C'$$
$endgroup$
– Mostafa Ayaz
Jan 5 at 4:16
$begingroup$
But x belongs to B
$endgroup$
– Nemanja Djordjevic
Jan 5 at 4:17
$begingroup$
But x belongs to B
$endgroup$
– Nemanja Djordjevic
Jan 5 at 4:17
$begingroup$
Sorry for the typo.........
$endgroup$
– Mostafa Ayaz
Jan 5 at 4:18
$begingroup$
Sorry for the typo.........
$endgroup$
– Mostafa Ayaz
Jan 5 at 4:18
1
1
$begingroup$
Yes because the taking complementary then intersection operation is associative i.e. $$((A-B_1)-B_2)-cdots - B_n=Abigcap_{i=1}^{n}B_i'=Acapleft(bigcup_{i=1}^{n}B_iright)'$$
$endgroup$
– Mostafa Ayaz
Jan 5 at 4:29
$begingroup$
Yes because the taking complementary then intersection operation is associative i.e. $$((A-B_1)-B_2)-cdots - B_n=Abigcap_{i=1}^{n}B_i'=Acapleft(bigcup_{i=1}^{n}B_iright)'$$
$endgroup$
– Mostafa Ayaz
Jan 5 at 4:29
|
show 3 more comments
$begingroup$
OK, we want to show
$$A Delta (B setminus C) = (A Delta B)setminus C Leftrightarrow A cap C = emptyset$$
Assume we have $A cap C neq emptyset$ (so trying for the contrapositive, which seems easier as it gives us something concrete to work with). Hence we have some $p in A cap C$. Case 1: $p in B$, then $p notin (B setminus C)$ (as $p in C$) but also $p in A$, so $p in A Delta (B setminus C)$ but on the other hand $p notin A Delta B$ so $p notin (A Delta B) setminus C$ so we have disproved the LHS identity in this case. Case 2: $p notin B$, then $p in A Delta B$ (and still $p in C$) so $p notin (A Delta B) setminus C$, and $p notin B setminus C$ and still $p in A$ so $p in A Delta (Bsetminus C)$ as in case 1, we have disproved the identity again.
Now assume that $A cap C = emptyset$ holds. We want to show the LHS identity which is really two inclusions. So suppose $x in A Delta (B setminus C)$, so case 1: $x in A$ and $x notin B setminus C$. As $x notin C$ (as otherwise $A cap C neq emptyset$) we know also that $x notin B$ (or else $x in B setminus C)$ after all) and so $x in (A Delta B)setminus C$, so the left to right inclusion holds for case 1. Case 2: $x notin A$ and $x in B setminus C$, and then $x in B$ and $x notin C$ and so $x in (A Delta B)setminus C$ and so the left to right inclusion also holds in case 2. So we're done with that inclusion. Now suppose that $x in (A Delta B) setminus C$ so $x in A Delta B$ and $x notin C$. Again 2 cases: case 1: $x in A$, $x notin B$, but then $x notin B setminus C$ and so $x in A Delta (B setminus C)$ and the right to left inclusion holds. Case 2: $x notin A$ and $x in B$, and then $x in B setminus C$ and so $x in A Delta (B setminus C)$ and the right to left inclusion holds (without even using $A cap C = emptyset$, so that inclusion always holds). So we have the LHS equality when $A$ and $C$ are disjoint.
answered Jan 5 at 6:43
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
add a comment |
$begingroup$
OK, we want to show
$$A Delta (B setminus C) = (A Delta B)setminus C Leftrightarrow A cap C = emptyset$$
Assume we have $A cap C neq emptyset$ (so trying for the contrapositive, which seems easier as it gives us something concrete to work with). Hence we have some $p in A cap C$. Case 1: $p in B$, then $p notin (B setminus C)$ (as $p in C$) but also $p in A$, so $p in A Delta (B setminus C)$ but on the other hand $p notin A Delta B$ so $p notin (A Delta B) setminus C$ so we have disproved the LHS identity in this case. Case 2: $p notin B$, then $p in A Delta B$ (and still $p in C$) so $p notin (A Delta B) setminus C$, and $p notin B setminus C$ and still $p in A$ so $p in A Delta (Bsetminus C)$ as in case 1, we have disproved the identity again.
Now assume that $A cap C = emptyset$ holds. We want to show the LHS identity which is really two inclusions. So suppose $x in A Delta (B setminus C)$, so case 1: $x in A$ and $x notin B setminus C$. As $x notin C$ (as otherwise $A cap C neq emptyset$) we know also that $x notin B$ (or else $x in B setminus C)$ after all) and so $x in (A Delta B)setminus C$, so the left to right inclusion holds for case 1. Case 2: $x notin A$ and $x in B setminus C$, and then $x in B$ and $x notin C$ and so $x in (A Delta B)setminus C$ and so the left to right inclusion also holds in case 2. So we're done with that inclusion. Now suppose that $x in (A Delta B) setminus C$ so $x in A Delta B$ and $x notin C$. Again 2 cases: case 1: $x in A$, $x notin B$, but then $x notin B setminus C$ and so $x in A Delta (B setminus C)$ and the right to left inclusion holds. Case 2: $x notin A$ and $x in B$, and then $x in B setminus C$ and so $x in A Delta (B setminus C)$ and the right to left inclusion holds (without even using $A cap C = emptyset$, so that inclusion always holds). So we have the LHS equality when $A$ and $C$ are disjoint.
answered Jan 5 at 6:43
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
add a comment |
$begingroup$
OK, we want to show
$$A Delta (B setminus C) = (A Delta B)setminus C Leftrightarrow A cap C = emptyset$$
Assume we have $A cap C neq emptyset$ (so trying for the contrapositive, which seems easier as it gives us something concrete to work with). Hence we have some $p in A cap C$. Case 1: $p in B$, then $p notin (B setminus C)$ (as $p in C$) but also $p in A$, so $p in A Delta (B setminus C)$ but on the other hand $p notin A Delta B$ so $p notin (A Delta B) setminus C$ so we have disproved the LHS identity in this case. Case 2: $p notin B$, then $p in A Delta B$ (and still $p in C$) so $p notin (A Delta B) setminus C$, and $p notin B setminus C$ and still $p in A$ so $p in A Delta (Bsetminus C)$ as in case 1, we have disproved the identity again.
Now assume that $A cap C = emptyset$ holds. We want to show the LHS identity which is really two inclusions. So suppose $x in A Delta (B setminus C)$, so case 1: $x in A$ and $x notin B setminus C$. As $x notin C$ (as otherwise $A cap C neq emptyset$) we know also that $x notin B$ (or else $x in B setminus C)$ after all) and so $x in (A Delta B)setminus C$, so the left to right inclusion holds for case 1. Case 2: $x notin A$ and $x in B setminus C$, and then $x in B$ and $x notin C$ and so $x in (A Delta B)setminus C$ and so the left to right inclusion also holds in case 2. So we're done with that inclusion. Now suppose that $x in (A Delta B) setminus C$ so $x in A Delta B$ and $x notin C$. Again 2 cases: case 1: $x in A$, $x notin B$, but then $x notin B setminus C$ and so $x in A Delta (B setminus C)$ and the right to left inclusion holds. Case 2: $x notin A$ and $x in B$, and then $x in B setminus C$ and so $x in A Delta (B setminus C)$ and the right to left inclusion holds (without even using $A cap C = emptyset$, so that inclusion always holds). So we have the LHS equality when $A$ and $C$ are disjoint.
answered Jan 5 at 6:43
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
OK, we want to show
$$A Delta (B setminus C) = (A Delta B)setminus C Leftrightarrow A cap C = emptyset$$
Assume we have $A cap C neq emptyset$ (so trying for the contrapositive, which seems easier as it gives us something concrete to work with). Hence we have some $p in A cap C$. Case 1: $p in B$, then $p notin (B setminus C)$ (as $p in C$) but also $p in A$, so $p in A Delta (B setminus C)$ but on the other hand $p notin A Delta B$ so $p notin (A Delta B) setminus C$ so we have disproved the LHS identity in this case. Case 2: $p notin B$, then $p in A Delta B$ (and still $p in C$) so $p notin (A Delta B) setminus C$, and $p notin B setminus C$ and still $p in A$ so $p in A Delta (Bsetminus C)$ as in case 1, we have disproved the identity again.
Now assume that $A cap C = emptyset$ holds. We want to show the LHS identity which is really two inclusions. So suppose $x in A Delta (B setminus C)$, so case 1: $x in A$ and $x notin B setminus C$. As $x notin C$ (as otherwise $A cap C neq emptyset$) we know also that $x notin B$ (or else $x in B setminus C)$ after all) and so $x in (A Delta B)setminus C$, so the left to right inclusion holds for case 1. Case 2: $x notin A$ and $x in B setminus C$, and then $x in B$ and $x notin C$ and so $x in (A Delta B)setminus C$ and so the left to right inclusion also holds in case 2. So we're done with that inclusion. Now suppose that $x in (A Delta B) setminus C$ so $x in A Delta B$ and $x notin C$. Again 2 cases: case 1: $x in A$, $x notin B$, but then $x notin B setminus C$ and so $x in A Delta (B setminus C)$ and the right to left inclusion holds. Case 2: $x notin A$ and $x in B$, and then $x in B setminus C$ and so $x in A Delta (B setminus C)$ and the right to left inclusion holds (without even using $A cap C = emptyset$, so that inclusion always holds). So we have the LHS equality when $A$ and $C$ are disjoint.
answered Jan 5 at 6:43
Henno BrandsmaHenno Brandsma
106k347114
answered Jan 5 at 6:43
Henno BrandsmaHenno Brandsma
106k347114
answered Jan 5 at 6:43
Henno BrandsmaHenno Brandsma
106k347114
answered Jan 5 at 6:43
Henno BrandsmaHenno Brandsma
106k347114
106k347114
add a comment |
add a comment |
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$begingroup$
Yes, that step is indeed wrong, because the set difference operation is not associative, so let's not do that. I'd suggest that you start by drawing Venn diagrams for both $Abigtriangleup(Bsetminus C)$ and $(Abigtriangleup B)setminus C$. You will see that, in general, they are not equal, as well as why $Acap C=varnothing$ is needed to make them equal. That may give you a hint how to approach the problem.
$endgroup$
– zipirovich
Jan 5 at 2:06
$begingroup$
I still don't know.
$endgroup$
– Nemanja Djordjevic
Jan 5 at 2:15
$begingroup$
Did you draw Venn diagrams for both expressions?
$endgroup$
– zipirovich
Jan 5 at 2:17
$begingroup$
yes and they are the same
$endgroup$
– Nemanja Djordjevic
Jan 5 at 2:35
$begingroup$
No, they are not.
$endgroup$
– zipirovich
Jan 5 at 2:36