
Is this proof correct?
I found a Abstract algebra book in goodwill (Rofl)
I can not get how do prove something. It seems almost impossible in some cases.
This is a question in it:
Prove :
Where
and U is the universal set.
Here is my attempted proof:
And the reverse is true. Therefore they must be equal.
Is this correct? (Bow)

Not all the steps are correct.
Your first step needs to be
Also,
implies
, not
That's essentially a DeMorgan Law application. You fix this error, incidentally, in the next step.
Also, you should be more careful with your parentheses here:
In context, I can tell that you meant
but you don't want to write so that people can understand. Write so that no one can misunderstand!
All your steps are reversible, correct? So just change your implications to double implications, and you can then omit the sentence "And the reverse is true." (You mean converse, by the way.)
As Strunk and White have told us, "Omit needless words!"

How does ? is the intersection so you would say A and B are true, else it is false. Thus logical conjunction :" "

Except that you are negating the whole thing. You're essentially saying this:
which is the same as
which is the same as
which is the same as
Does that make sense?

Hi, I will try prove this using pure logic notation.
To prove we need to show the following to be true:
1. If then
and
2. If then
Now I will show you how to prove 1. You can try 2. yourself (Smirk)
Let and now 1. becomes
To prove this conditional proposition is true we will now go through all the possibilities. If p is false, then is vacuously true. We can ignore this trivial case.
Thus we need to show if p is true and q is true then is true.
Assume p is true.
Now p is only true under 3 conditions: (using the 'inclusiveor' definition we have)
a) is true and is false
b) is false and is true
c) is true and is true
Now notice in a)
So if is true then is false. Similarly is true.
Likewise from b) you should conclude is true and is false.
c) is false and is false.
Now consider
Using the results from a), b) and c) we see q is always true. Thus 1. is now proven.
Try 2. with the same principles :)

Oh! I get it!.
I was thinking of OR as XOR.
If we use OR then the elements could be in the location AND is.
Proving that ( A intercect B)' is a subset. And once we prove the converse, we prove they are equal.
Right?

I'm not sure I follow this post. Are you referring to usagi_killer's proof in post # 5?