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Prove that: (A U B)^c - C^c = A^c - (B U C)^c

  • Thread starter steveT
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  • #1
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Homework Statement


I need to prove that: (A U B)^c - C^c = A^c - (B U C)^c



Homework Equations





The Attempt at a Solution



I know that (A U B)^c = A^c and B^c

My problem is I'm not sure how to rearrange or distribute the minus sign to make it equal to the other side.

Thanks for the help
 

Answers and Replies

  • #2
Dick
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Homework Statement


I need to prove that: (A U B)^c - C^c = A^c - (B U C)^c



Homework Equations





The Attempt at a Solution



I know that (A U B)^c = A^c and B^c

My problem is I'm not sure how to rearrange or distribute the minus sign to make it equal to the other side.

Thanks for the help
Your expression is generally false. Pick A={1}, B={2} and C={3} all subsets of {1,2,3}. You probably have a typo in the statement. In general for problems like this you can use A-B=AnB^C and use deMorgan.
 
  • #3
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What part is generally false? Maybe this is more clear.

The complement of ( A U B) - The complement of C = The complement of A - The complement of (B U C)

I know that the complement of ( A U B ) = the complement of A and the complement of B
 
  • #4
Dick
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What part is generally false? Maybe this is more clear.

The complement of ( A U B) - The complement of C = The complement of A - The complement of (B U C)

I know that the complement of ( A U B ) = the complement of A and the complement of B
That's clear enough. But take the example I gave you in post 2 and put it into that statement. Are the right and left sides equal?
 
  • #5
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Wow, I just wasted a lot of time, I wrote it down wrong. It should have been this

(A U B)^c - C^c = A^c - (B U C^c)

Ok, now that that is settled, I'm trying to prove the RHS first given the LHS.

Like I stated before, I know that

(A U B)^c = A^c and B^c

So I get

A^c and B^c - C^c

But still not sure how to rearrange the minus sign, is it something like this

A^c - B^c and C^c =
A^c - (B U C^c)

Or am I completely off
 
  • #6
Dick
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The difference of two sets A-B is the same as AnB^c. Use that to get rid of the minus signs and turn everything into intersections and unions.
 
  • #7
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OK, here is what I got so far, still trying to prove the RHS

(A U B)^c - C^c

A^c n B^c - C^c

Since A^c n B^c = A^c - B

I get

A^c - B - C^c

Now I'm stuck again
 
  • #8
Dick
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OK, here is what I got so far, still trying to prove the RHS

(A U B)^c - C^c

A^c n B^c - C^c

Since A^c n B^c = A^c - B

I get

A^c - B - C^c

Now I'm stuck again
I suggested you change everything into intersections and unions. Instead you changed everything into differences (minuses). Try the other way around.
 
  • #9
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How about this

(A U B)^c - C^c = A^c n B^c n C

Still not sure how to make that look like the RHS of the equation
 
  • #10
Dick
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How about this

(A U B)^c - C^c = A^c n B^c n C

Still not sure how to make that look like the RHS of the equation
Change the RHS into intersections and unions too. Then see if they look the same.
 
  • #11
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(A U B)^c - C^c = A^c - (B U C^c)

A^c n B^c n C = (A^c n B^c) n (A^c n C)

A^c n B^c n C = (A U B)^c n (A^c n C)

A^c n B^c n C = (A U B)^c n (A^c - C^c)

I'm getting close, is this correct so far. Not sure where to go now?
 
  • #12
Dick
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(A U B)^c - C^c = A^c - (B U C^c)

A^c n B^c n C = (A^c n B^c) n (A^c n C)

A^c n B^c n C = (A U B)^c n (A^c n C)

A^c n B^c n C = (A U B)^c n (A^c - C^c)

I'm getting close, is this correct so far. Not sure where to go now?
You overshot. You told me the LHS was the same as A^c n B^c n C in post 9. Look at the second line above.
 
  • #13
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On the LHS I'm stuck at A^c n B^c n C, I'm trying to get it to equal the original RHS. As far as everything I did on the RHS, I was trying to get it to equal the LHS
 
  • #14
Dick
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On the LHS I'm stuck at A^c n B^c n C, I'm trying to get it to equal the original RHS. As far as everything I did on the RHS, I was trying to get it to equal the LHS
If you have both the LHS and the RHS equal to A^c n B^c n C then you are done. They are equal. All of your steps are reversible.
 
  • #15
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If you have both the LHS and the RHS equal to A^c n B^c n C then you are done. They are equal. All of your steps are reversible.
But I don't have the LHS = RHS anywhere. For a proof like this I need to get the original LHS to look like the original RHS, then I need to get the original RHS to look like the original LHS. I have done neither.
 
  • #16
Dick
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But I don't have the LHS = RHS anywhere. For a proof like this I need to get the original LHS to look like the original RHS, then I need to get the original RHS to look like the original LHS. I have done neither.
If you feel that compulsive about it then do that. Turn the LHS into A^c n B^c n C and then turn A^c n B^c n C into the RHS. You've got all the steps already.
 

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