Set Theory: Proving A-(BUC)=(A\cup B)-C

Carmen12
Messages
8
Reaction score
0

Homework Statement





[A-(BUC)]U[(A\cup B)-C]U[(A\cap C)-B]U[A\capB\capC];

The Attempt at a Solution



Sorry about the crappy formatting (btw).
Anyway, I'm trying to "prove" that this is is equal to A. So basically cancelling out the Bs and Cs? I'm not sure how to go about this. de morgan's laws? *sigh* I tested it with Venn Diagrams and it is equal to A.
 
Last edited:
Physics news on Phys.org
I would start be re-writing the subtractions a little more explicitly, as an example
A - B = A - A \cap B

this should lead to a few siplifications, and you should be able to apply de morgan's law more easily
 
Thank you so much! Some research on de morgan is in need!

I wish there was a calculator for this stuff. *sigh*

For the proof will just showing the calculations work? Or is there a particular way of writing it out I need?
 
just try and simplify the RHS as much as you can

and its not De Morgan's law you need, which deals with complements, its just distrbutivity & associtivity see the following

http://en.wikipedia.org/wiki/Algebra_of_sets
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top