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can someone help me prove this
Can someone help me solve this misunderstanding please.Can someone help me solve this problem please.Can someone explain this proof by case example?Can someone explain me this summation?Can someone help me derive this equation using Euler's formula?Can someone help me with this question about relationships?Can someone help me with this relationship thinking?can someone help me understand this solution (introductory discrete math)Can you help to prove this equation by inductionCan someone help me simplify this boolean algebra
$begingroup$
$$ 2^{(AΔB)}={S|S = Xcup Y} $$ where $ X∈ 2^A, Y∈ 2^B$, and $$S= Zcup W $$ where $Z∈ 2^{(A)^c}, W∈2^{(B)^c}$
discrete-mathematics
edited Mar 20 at 19:23
Mark Fischler
34k12552
asked Mar 20 at 19:18
Mahmood JazmawyMahmood Jazmawy
143
$endgroup$
|
show 1 more comment
$begingroup$
$$ 2^{(AΔB)}={S|S = Xcup Y} $$ where $ X∈ 2^A, Y∈ 2^B$, and $$S= Zcup W $$ where $Z∈ 2^{(A)^c}, W∈2^{(B)^c}$
discrete-mathematics
edited Mar 20 at 19:23
Mark Fischler
34k12552
asked Mar 20 at 19:18
Mahmood JazmawyMahmood Jazmawy
143
$endgroup$
$begingroup$
What is the meaning of $Delta$ in the set $ADelta B$?
$endgroup$
– Mark Fischler
Mar 20 at 19:25
$begingroup$
@MarkFischler (Acup B) - (Acap B)
$endgroup$
– Mahmood Jazmawy
Mar 20 at 19:26
$begingroup$
symmetric difference
$endgroup$
– ADITYA PRAKASH
Mar 20 at 19:28
$begingroup$
@Mahmood Jazmawy your question seems to be wrong , $S=X cup Y -(X cap Y)$ , according to me this should be equivalent to $2^{A Delta B}$ , should this be true , then I can post an answer , please verify.
$endgroup$
– ADITYA PRAKASH
Mar 20 at 19:30
$begingroup$
@ADITYAPRAKASH no the question as is I didn't change anything in it and the 2AΔB is as I said
$endgroup$
– Mahmood Jazmawy
Mar 20 at 19:35
|
show 1 more comment
$begingroup$
$$ 2^{(AΔB)}={S|S = Xcup Y} $$ where $ X∈ 2^A, Y∈ 2^B$, and $$S= Zcup W $$ where $Z∈ 2^{(A)^c}, W∈2^{(B)^c}$
discrete-mathematics
edited Mar 20 at 19:23
Mark Fischler
34k12552
asked Mar 20 at 19:18
Mahmood JazmawyMahmood Jazmawy
143
$endgroup$
$$ 2^{(AΔB)}={S|S = Xcup Y} $$ where $ X∈ 2^A, Y∈ 2^B$, and $$S= Zcup W $$ where $Z∈ 2^{(A)^c}, W∈2^{(B)^c}$
discrete-mathematics
discrete-mathematics
edited Mar 20 at 19:23
Mark Fischler
34k12552
asked Mar 20 at 19:18
Mahmood JazmawyMahmood Jazmawy
143
edited Mar 20 at 19:23
Mark Fischler
34k12552
asked Mar 20 at 19:18
Mahmood JazmawyMahmood Jazmawy
143
edited Mar 20 at 19:23
Mark Fischler
34k12552
edited Mar 20 at 19:23
Mark Fischler
34k12552
edited Mar 20 at 19:23
Mark Fischler
34k12552
34k12552
asked Mar 20 at 19:18
Mahmood JazmawyMahmood Jazmawy
143
asked Mar 20 at 19:18
Mahmood JazmawyMahmood Jazmawy
143
asked Mar 20 at 19:18
Mahmood JazmawyMahmood Jazmawy
143
143
$begingroup$
What is the meaning of $Delta$ in the set $ADelta B$?
$endgroup$
– Mark Fischler
Mar 20 at 19:25
$begingroup$
@MarkFischler (Acup B) - (Acap B)
$endgroup$
– Mahmood Jazmawy
Mar 20 at 19:26
$begingroup$
symmetric difference
$endgroup$
– ADITYA PRAKASH
Mar 20 at 19:28
$begingroup$
@Mahmood Jazmawy your question seems to be wrong , $S=X cup Y -(X cap Y)$ , according to me this should be equivalent to $2^{A Delta B}$ , should this be true , then I can post an answer , please verify.
$endgroup$
– ADITYA PRAKASH
Mar 20 at 19:30
$begingroup$
@ADITYAPRAKASH no the question as is I didn't change anything in it and the 2AΔB is as I said
$endgroup$
– Mahmood Jazmawy
Mar 20 at 19:35
|
show 1 more comment
$begingroup$
What is the meaning of $Delta$ in the set $ADelta B$?
$endgroup$
– Mark Fischler
Mar 20 at 19:25
$begingroup$
@MarkFischler (Acup B) - (Acap B)
$endgroup$
– Mahmood Jazmawy
Mar 20 at 19:26
$begingroup$
symmetric difference
$endgroup$
– ADITYA PRAKASH
Mar 20 at 19:28
$begingroup$
@Mahmood Jazmawy your question seems to be wrong , $S=X cup Y -(X cap Y)$ , according to me this should be equivalent to $2^{A Delta B}$ , should this be true , then I can post an answer , please verify.
$endgroup$
– ADITYA PRAKASH
Mar 20 at 19:30
$begingroup$
@ADITYAPRAKASH no the question as is I didn't change anything in it and the 2AΔB is as I said
$endgroup$
– Mahmood Jazmawy
Mar 20 at 19:35
$begingroup$
What is the meaning of $Delta$ in the set $ADelta B$?
$endgroup$
– Mark Fischler
Mar 20 at 19:25
$begingroup$
What is the meaning of $Delta$ in the set $ADelta B$?
$endgroup$
– Mark Fischler
Mar 20 at 19:25
$begingroup$
@MarkFischler (Acup B) - (Acap B)
$endgroup$
– Mahmood Jazmawy
Mar 20 at 19:26
$begingroup$
@MarkFischler (Acup B) - (Acap B)
$endgroup$
– Mahmood Jazmawy
Mar 20 at 19:26
$begingroup$
symmetric difference
$endgroup$
– ADITYA PRAKASH
Mar 20 at 19:28
$begingroup$
symmetric difference
$endgroup$
– ADITYA PRAKASH
Mar 20 at 19:28
$begingroup$
@Mahmood Jazmawy your question seems to be wrong , $S=X cup Y -(X cap Y)$ , according to me this should be equivalent to $2^{A Delta B}$ , should this be true , then I can post an answer , please verify.
$endgroup$
– ADITYA PRAKASH
Mar 20 at 19:30
$begingroup$
@Mahmood Jazmawy your question seems to be wrong , $S=X cup Y -(X cap Y)$ , according to me this should be equivalent to $2^{A Delta B}$ , should this be true , then I can post an answer , please verify.
$endgroup$
– ADITYA PRAKASH
Mar 20 at 19:30
$begingroup$
@ADITYAPRAKASH no the question as is I didn't change anything in it and the 2AΔB is as I said
$endgroup$
– Mahmood Jazmawy
Mar 20 at 19:35
$begingroup$
@ADITYAPRAKASH no the question as is I didn't change anything in it and the 2AΔB is as I said
$endgroup$
– Mahmood Jazmawy
Mar 20 at 19:35
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
I thought a diagram will be useful here.As you can see I have drawn sets A & B and X & Y.The symmetric difference of A and B i.e. $ADelta B$ is coloured in buff color(light orange).You have defined the process of moving from set $Acup B$ to $Xcup Y$ as raising 2 to the power of elements from the former to the latter.It's evident that you cannot select such a power of 2 from the set $Acup B$ that lies in $Acap B$.It implies seamlessly that the valid elements in set S won't fall in $Xcap Y$ because you are not able to select elements from $Acap B$ in the first place and thus $$2^{(ADelta B)}={S|S=(Xcup Y)-(Xcap Y)}$$. This is the first part of your answer , You can do the second part in a similar fashion.
answered Mar 20 at 19:58
ADITYA PRAKASHADITYA PRAKASH
365110
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
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oldest
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votes
$begingroup$
I thought a diagram will be useful here.As you can see I have drawn sets A & B and X & Y.The symmetric difference of A and B i.e. $ADelta B$ is coloured in buff color(light orange).You have defined the process of moving from set $Acup B$ to $Xcup Y$ as raising 2 to the power of elements from the former to the latter.It's evident that you cannot select such a power of 2 from the set $Acup B$ that lies in $Acap B$.It implies seamlessly that the valid elements in set S won't fall in $Xcap Y$ because you are not able to select elements from $Acap B$ in the first place and thus $$2^{(ADelta B)}={S|S=(Xcup Y)-(Xcap Y)}$$. This is the first part of your answer , You can do the second part in a similar fashion.
answered Mar 20 at 19:58
ADITYA PRAKASHADITYA PRAKASH
365110
$endgroup$
add a comment |
$begingroup$
I thought a diagram will be useful here.As you can see I have drawn sets A & B and X & Y.The symmetric difference of A and B i.e. $ADelta B$ is coloured in buff color(light orange).You have defined the process of moving from set $Acup B$ to $Xcup Y$ as raising 2 to the power of elements from the former to the latter.It's evident that you cannot select such a power of 2 from the set $Acup B$ that lies in $Acap B$.It implies seamlessly that the valid elements in set S won't fall in $Xcap Y$ because you are not able to select elements from $Acap B$ in the first place and thus $$2^{(ADelta B)}={S|S=(Xcup Y)-(Xcap Y)}$$. This is the first part of your answer , You can do the second part in a similar fashion.
answered Mar 20 at 19:58
ADITYA PRAKASHADITYA PRAKASH
365110
$endgroup$
add a comment |
$begingroup$
I thought a diagram will be useful here.As you can see I have drawn sets A & B and X & Y.The symmetric difference of A and B i.e. $ADelta B$ is coloured in buff color(light orange).You have defined the process of moving from set $Acup B$ to $Xcup Y$ as raising 2 to the power of elements from the former to the latter.It's evident that you cannot select such a power of 2 from the set $Acup B$ that lies in $Acap B$.It implies seamlessly that the valid elements in set S won't fall in $Xcap Y$ because you are not able to select elements from $Acap B$ in the first place and thus $$2^{(ADelta B)}={S|S=(Xcup Y)-(Xcap Y)}$$. This is the first part of your answer , You can do the second part in a similar fashion.
answered Mar 20 at 19:58
ADITYA PRAKASHADITYA PRAKASH
365110
$endgroup$
I thought a diagram will be useful here.As you can see I have drawn sets A & B and X & Y.The symmetric difference of A and B i.e. $ADelta B$ is coloured in buff color(light orange).You have defined the process of moving from set $Acup B$ to $Xcup Y$ as raising 2 to the power of elements from the former to the latter.It's evident that you cannot select such a power of 2 from the set $Acup B$ that lies in $Acap B$.It implies seamlessly that the valid elements in set S won't fall in $Xcap Y$ because you are not able to select elements from $Acap B$ in the first place and thus $$2^{(ADelta B)}={S|S=(Xcup Y)-(Xcap Y)}$$. This is the first part of your answer , You can do the second part in a similar fashion.
answered Mar 20 at 19:58
ADITYA PRAKASHADITYA PRAKASH
365110
answered Mar 20 at 19:58
ADITYA PRAKASHADITYA PRAKASH
365110
answered Mar 20 at 19:58
ADITYA PRAKASHADITYA PRAKASH
365110
answered Mar 20 at 19:58
ADITYA PRAKASHADITYA PRAKASH
365110
365110
add a comment |
add a comment |
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$begingroup$
What is the meaning of $Delta$ in the set $ADelta B$?
$endgroup$
– Mark Fischler
Mar 20 at 19:25
$begingroup$
@MarkFischler (Acup B) - (Acap B)
$endgroup$
– Mahmood Jazmawy
Mar 20 at 19:26
$begingroup$
symmetric difference
$endgroup$
– ADITYA PRAKASH
Mar 20 at 19:28
$begingroup$
@Mahmood Jazmawy your question seems to be wrong , $S=X cup Y -(X cap Y)$ , according to me this should be equivalent to $2^{A Delta B}$ , should this be true , then I can post an answer , please verify.
$endgroup$
– ADITYA PRAKASH
Mar 20 at 19:30
$begingroup$
@ADITYAPRAKASH no the question as is I didn't change anything in it and the 2AΔB is as I said
$endgroup$
– Mahmood Jazmawy
Mar 20 at 19:35