Set Theory Proofing: Prove that (A ^ B)' ^ B = B ^ A'

MathMind

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Apologies if this is the incorrect forum for this question..

I would really appreciate some help in proofing the following.

(A ^ B)' ^ B = B ^ A'

Reads as complement of A intersect B , intersected with B, is equal to B intersect A complement.
 
Apologies if this is the incorrect forum for this question..

I would really appreciate some help in proofing the following.

(A ^ B)' ^ B = B ^ A'

Reads as complement of A intersect B , intersected with B, is equal to B intersect A complement.

What are your thoughts?

Please share your work with us ...even if you know it is wrong

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You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

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Apologies if this is the incorrect forum for this question..
I would really appreciate some help in proofing the following.
\(\displaystyle (A \cap B)^c\cap B = B \cap A^c\)
Reads as complement of A intersect B , intersected with B, is equal to B intersect A complement.
See the quote above. It is a lot more readable is it not?
Here is the code [ tex](A \cap B)^c\cap B = B \cap A^c[/tex](without the first space)

Start with \(\displaystyle (A \cap B)^c = (A^c \cup B^c)\cap B\). Use distribution
 
I would really appreciate some help in proofing the following.

(A ^ B)' ^ B = B ^ A'

Reads as complement of A intersect B , intersected with B, is equal to B intersect A complement.
Try using the standard method of "element-chasing".

Pick a typical, generic element of, say, the left-hand side of the proposed equality. Can you should that this element must necessarily also be an element of the right-hand side? Then you've proved that the left-hand side is a subset of the right-hand side. Then go the opposite direction.

Once you've shown "double set inclusion" (that is, that each set is a subset of the other), then you've proved that the sets must be the same, and are thus equal.

If you get stuck, please reply with a clear listing of your thoughts and steps so far. Thank you! ;)
 
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