Simplifying $(A-B^c)∪(B∩(A∩B)^c)$
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I'm looking to use laws of set algebra to simplify:
$(A-B^c)∪(B∩(A∩B)^c)$
But I'm currently stuck on what I should do.
= $(A-B^c)∪(B∩(A^c∪B^c))$ (De Morgan's)
= $(A-B^c)∪((B∩A^c)∪(B∩B^c))$ (Distributive Law)
= $(A-B^c)∪(B∩A^c)$ (Intersection with Compliment)
And from here, I'm stuck. If I draw up a Venn Diagram, I can see that the simplified expression would just be $B$, but I'm currently stuck on how to get there. I've tried turning $(B∩A^c)$ to $B - A$, but that didn't seem to help me much.
discrete-mathematics elementary-set-theory
asked Aug 14 at 6:48
hdnmages
61
add a comment |Â
up vote
1
down vote
favorite
I'm looking to use laws of set algebra to simplify:
$(A-B^c)∪(B∩(A∩B)^c)$
But I'm currently stuck on what I should do.
= $(A-B^c)∪(B∩(A^c∪B^c))$ (De Morgan's)
= $(A-B^c)∪((B∩A^c)∪(B∩B^c))$ (Distributive Law)
= $(A-B^c)∪(B∩A^c)$ (Intersection with Compliment)
And from here, I'm stuck. If I draw up a Venn Diagram, I can see that the simplified expression would just be $B$, but I'm currently stuck on how to get there. I've tried turning $(B∩A^c)$ to $B - A$, but that didn't seem to help me much.
discrete-mathematics elementary-set-theory
asked Aug 14 at 6:48
hdnmages
61
Are $A, Bsubseteq X$?
– Cornman
Aug 14 at 6:49
Hi, the question doesn't specify any specific relations for A and B, so I'm not sure.
– hdnmages
Aug 14 at 6:51
$A-B^c=A cap B$
– nicomezi
Aug 14 at 6:52
Thanks. I completely missed that, and can now apply distributive law which helped me finish the question.
– hdnmages
Aug 14 at 6:56
Well, but where do you build the complement of $B$?
– Cornman
Aug 14 at 7:08
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm looking to use laws of set algebra to simplify:
$(A-B^c)∪(B∩(A∩B)^c)$
But I'm currently stuck on what I should do.
= $(A-B^c)∪(B∩(A^c∪B^c))$ (De Morgan's)
= $(A-B^c)∪((B∩A^c)∪(B∩B^c))$ (Distributive Law)
= $(A-B^c)∪(B∩A^c)$ (Intersection with Compliment)
And from here, I'm stuck. If I draw up a Venn Diagram, I can see that the simplified expression would just be $B$, but I'm currently stuck on how to get there. I've tried turning $(B∩A^c)$ to $B - A$, but that didn't seem to help me much.
discrete-mathematics elementary-set-theory
asked Aug 14 at 6:48
hdnmages
61
I'm looking to use laws of set algebra to simplify:
$(A-B^c)∪(B∩(A∩B)^c)$
But I'm currently stuck on what I should do.
= $(A-B^c)∪(B∩(A^c∪B^c))$ (De Morgan's)
= $(A-B^c)∪((B∩A^c)∪(B∩B^c))$ (Distributive Law)
= $(A-B^c)∪(B∩A^c)$ (Intersection with Compliment)
And from here, I'm stuck. If I draw up a Venn Diagram, I can see that the simplified expression would just be $B$, but I'm currently stuck on how to get there. I've tried turning $(B∩A^c)$ to $B - A$, but that didn't seem to help me much.
discrete-mathematics elementary-set-theory
asked Aug 14 at 6:48
hdnmages
61
asked Aug 14 at 6:48
hdnmages
61
asked Aug 14 at 6:48
hdnmages
61
asked Aug 14 at 6:48
hdnmages
61
61
Are $A, Bsubseteq X$?
– Cornman
Aug 14 at 6:49
Hi, the question doesn't specify any specific relations for A and B, so I'm not sure.
– hdnmages
Aug 14 at 6:51
$A-B^c=A cap B$
– nicomezi
Aug 14 at 6:52
Thanks. I completely missed that, and can now apply distributive law which helped me finish the question.
– hdnmages
Aug 14 at 6:56
Well, but where do you build the complement of $B$?
– Cornman
Aug 14 at 7:08
add a comment |Â
Are $A, Bsubseteq X$?
– Cornman
Aug 14 at 6:49
Hi, the question doesn't specify any specific relations for A and B, so I'm not sure.
– hdnmages
Aug 14 at 6:51
$A-B^c=A cap B$
– nicomezi
Aug 14 at 6:52
Thanks. I completely missed that, and can now apply distributive law which helped me finish the question.
– hdnmages
Aug 14 at 6:56
Well, but where do you build the complement of $B$?
– Cornman
Aug 14 at 7:08
Are $A, Bsubseteq X$?
– Cornman
Aug 14 at 6:49
Are $A, Bsubseteq X$?
– Cornman
Aug 14 at 6:49
Hi, the question doesn't specify any specific relations for A and B, so I'm not sure.
– hdnmages
Aug 14 at 6:51
Hi, the question doesn't specify any specific relations for A and B, so I'm not sure.
– hdnmages
Aug 14 at 6:51
$A-B^c=A cap B$
– nicomezi
Aug 14 at 6:52
$A-B^c=A cap B$
– nicomezi
Aug 14 at 6:52
Thanks. I completely missed that, and can now apply distributive law which helped me finish the question.
– hdnmages
Aug 14 at 6:56
Thanks. I completely missed that, and can now apply distributive law which helped me finish the question.
– hdnmages
Aug 14 at 6:56
Well, but where do you build the complement of $B$?
– Cornman
Aug 14 at 7:08
Well, but where do you build the complement of $B$?
– Cornman
Aug 14 at 7:08
add a comment |Â
2 Answers
2
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2
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We have $$(A-B^c)cup(Bcap A^c)=(Acap B)cup(B,,text A)=B$$
answered Aug 14 at 6:56
TheSimpliFire
10.1k61953
add a comment |Â
up vote
0
down vote
$(A−B^c)∪(B∩(A∩B)^c)$
First go with De-Morgan's:
$B ∩ (A ∩ B)^c$ = $B ∩ (A^c ∪ B^c)$
Applying Distributive Law, $B ∩ (A^c ∪ B^c)$ = $(A^c ∩ B) ∪ (B ∩ B^c)$ = $(A^c ∩ B)$ - (1)
As, $A - B$ = $A ∩ B^c$. Therfore, $A - B^c$ = $A ∩ B$ -(2)
Using (1)&(2),
$(A ∩ B) ∪ (A^c ∩ B)$ = $(A ∩ A^c) ∪ B$ = $B$
answered Aug 14 at 7:19
bhatejaud
84
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
We have $$(A-B^c)cup(Bcap A^c)=(Acap B)cup(B,,text A)=B$$
answered Aug 14 at 6:56
TheSimpliFire
10.1k61953
add a comment |Â
up vote
2
down vote
We have $$(A-B^c)cup(Bcap A^c)=(Acap B)cup(B,,text A)=B$$
answered Aug 14 at 6:56
TheSimpliFire
10.1k61953
add a comment |Â
up vote
2
down vote
up vote
2
down vote
We have $$(A-B^c)cup(Bcap A^c)=(Acap B)cup(B,,text A)=B$$
answered Aug 14 at 6:56
TheSimpliFire
10.1k61953
We have $$(A-B^c)cup(Bcap A^c)=(Acap B)cup(B,,text A)=B$$
answered Aug 14 at 6:56
TheSimpliFire
10.1k61953
answered Aug 14 at 6:56
TheSimpliFire
10.1k61953
answered Aug 14 at 6:56
TheSimpliFire
10.1k61953
answered Aug 14 at 6:56
TheSimpliFire
10.1k61953
10.1k61953
add a comment |Â
add a comment |Â
up vote
0
down vote
$(A−B^c)∪(B∩(A∩B)^c)$
First go with De-Morgan's:
$B ∩ (A ∩ B)^c$ = $B ∩ (A^c ∪ B^c)$
Applying Distributive Law, $B ∩ (A^c ∪ B^c)$ = $(A^c ∩ B) ∪ (B ∩ B^c)$ = $(A^c ∩ B)$ - (1)
As, $A - B$ = $A ∩ B^c$. Therfore, $A - B^c$ = $A ∩ B$ -(2)
Using (1)&(2),
$(A ∩ B) ∪ (A^c ∩ B)$ = $(A ∩ A^c) ∪ B$ = $B$
answered Aug 14 at 7:19
bhatejaud
84
add a comment |Â
up vote
0
down vote
$(A−B^c)∪(B∩(A∩B)^c)$
First go with De-Morgan's:
$B ∩ (A ∩ B)^c$ = $B ∩ (A^c ∪ B^c)$
Applying Distributive Law, $B ∩ (A^c ∪ B^c)$ = $(A^c ∩ B) ∪ (B ∩ B^c)$ = $(A^c ∩ B)$ - (1)
As, $A - B$ = $A ∩ B^c$. Therfore, $A - B^c$ = $A ∩ B$ -(2)
Using (1)&(2),
$(A ∩ B) ∪ (A^c ∩ B)$ = $(A ∩ A^c) ∪ B$ = $B$
answered Aug 14 at 7:19
bhatejaud
84
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$(A−B^c)∪(B∩(A∩B)^c)$
First go with De-Morgan's:
$B ∩ (A ∩ B)^c$ = $B ∩ (A^c ∪ B^c)$
Applying Distributive Law, $B ∩ (A^c ∪ B^c)$ = $(A^c ∩ B) ∪ (B ∩ B^c)$ = $(A^c ∩ B)$ - (1)
As, $A - B$ = $A ∩ B^c$. Therfore, $A - B^c$ = $A ∩ B$ -(2)
Using (1)&(2),
$(A ∩ B) ∪ (A^c ∩ B)$ = $(A ∩ A^c) ∪ B$ = $B$
answered Aug 14 at 7:19
bhatejaud
84
$(A−B^c)∪(B∩(A∩B)^c)$
First go with De-Morgan's:
$B ∩ (A ∩ B)^c$ = $B ∩ (A^c ∪ B^c)$
Applying Distributive Law, $B ∩ (A^c ∪ B^c)$ = $(A^c ∩ B) ∪ (B ∩ B^c)$ = $(A^c ∩ B)$ - (1)
As, $A - B$ = $A ∩ B^c$. Therfore, $A - B^c$ = $A ∩ B$ -(2)
Using (1)&(2),
$(A ∩ B) ∪ (A^c ∩ B)$ = $(A ∩ A^c) ∪ B$ = $B$
answered Aug 14 at 7:19
bhatejaud
84
answered Aug 14 at 7:19
bhatejaud
84
answered Aug 14 at 7:19
bhatejaud
84
answered Aug 14 at 7:19
bhatejaud
84
84
add a comment |Â
add a comment |Â
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Are $A, Bsubseteq X$?
– Cornman
Aug 14 at 6:49
Hi, the question doesn't specify any specific relations for A and B, so I'm not sure.
– hdnmages
Aug 14 at 6:51
$A-B^c=A cap B$
– nicomezi
Aug 14 at 6:52
Thanks. I completely missed that, and can now apply distributive law which helped me finish the question.
– hdnmages
Aug 14 at 6:56
Well, but where do you build the complement of $B$?
– Cornman
Aug 14 at 7:08