Question on Discrete math for CS on C about sets need help o

Question on Discrete math for CS on C++ about sets, need help on this part

1)Pick one law of sets and give an every application of it..

2) Do you think it is useful to personally know?

Solution

1. De Morgan\'s Laws

De Morgan\'s Laws, which tell us how to get rid of brackets, when there is a complement symbol () outside the brackets. This laws is:-

(A B) = A B

Look closely at how this Law works. The complement symbol after the bracket affects all three symbols inside the bracket when the brackets are removed.

As you can see A becomes A , B becomes B and becomes .

To prove this law, note first of all that when we defined a subset we said that if

A B and B A, then A = B

So we prove:

(i) (A B) A B

and then the other way round:

(ii) A B (A B)

The proof of (i) goes like this:

Let\'s pick an element at random x (A B) .

x could be a number, a function or anything. All we do know about x, is that x (A B)

So x (A B) because that\'s what complement means.

This means that x answers No to both questions Are you in A? and Are you in B? (otherwise it would be in the union of A and B). Therefore x A and x B

Applying complements again we get x A and x B

Finally, if something is in two sets, it must be in their intersection, so x A B

So, any element we pick at random from (A B) is definitely in A B . So by definition (A B) A B

The proof of (ii) is similar:

First, we pick an element at random from the first set, x A B

Using what we know about intersections, that means x A and x B

So, using what we know about complements, x A and x B

And if something is in neither A nor B, it can\'t be in their union, so x A B

So, finally: x (A B)

So: A B (A B)

We\'ve now proved (i) and (ii), and therefore: (A B) = A B

2. De Morgans law is very useful in solving complex digital circuits into simpler one. Hence we can design a complex circuit with simple devices like OR GATE and AND GATE.