# Why is my proof of this set identity incorrect?

• ainster31
It's not like multiplying or dividing both sides of equations in algebra.In summary, the conversation discusses the differences between regular algebra and set algebra, particularly in terms of reversible steps. While reversible steps are possible in algebra, they are not always applicable in set algebra, as certain equations and operations do not necessarily lead to reversible conclusions. The conversation also touches on the importance of starting with the given information and using set algebra rules to work towards the desired solution.
ainster31

## The Attempt at a Solution

$$A-(A\cap B)=A-B\\ A\cap (A\cap B)^{ C }=A\cap B^{ C }\quad (set\quad difference\quad law)\\ A\cup [A\cap (A\cap B)^{ C }]=A\cup [A\cap B^{ C }]\quad (applied\quad A\cup \quad to\quad both\quad sides)\\ A=A\quad (absorption \quad law)$$

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Among other things, it looks like you have started with what you wanted to prove and drew the conclusion that A = A. I'm guessing that the problem wasn't to prove that A = A.

Also, let's take an example where A and B are any two subsets if U. Here's a proof that A = B, using your argument:

##A = B##. Union U with both sides: ##A\cup U = B\cup U##. Therefore ##U = U##. End of proof.

LCKurtz said:
Among other things, it looks like you have started with what you wanted to prove and drew the conclusion that A = A. I'm guessing that the problem wasn't to prove that A = A.

Well, I can just reverse the steps to fix that.

LCKurtz said:
Also, let's take an example where A and B are any two subsets if U. Here's a proof that A = B, using your argument:

##A = B##. Union U with both sides: ##A\cup U = B\cup U##. Therefore ##U = U##. End of proof.

Hmm... How come in algebra I can do operations on both sides but I can't do that in set algebra?

For example in algebra, I can do this:

4 = 4
4 + 5 = 4 + 5
9 = 9

But I can't seem to do that with set algebra?

ainster31 said:
Well, I can just reverse the steps to fix that.

Hmm... How come in algebra I can do operations on both sides but I can't do that in set algebra?

For example in algebra, I can do this:

4 = 4
4 + 5 = 4 + 5
9 = 9

But I can't seem to do that with set algebra?

That's right. You can't because it doesn't work that way. And even in ordinary algebra, you don't start with something you are trying to prove by assuming it is true. Then you would get arguments like this: To prove 3 = 5 start with 3 = 5. Multiply both sides by 0 and get 0 = 0. That doesn't prove 3 = 5.

In your current problem you could start with the left side and use your rules on it. You already have ##A - (A\cap B) = A \cap (A\cap B)^c##. Just keep going with it using the rules until you get an expression identical to the right side of the equation.

LCKurtz said:
That's right. You can't because it doesn't work that way. And even in ordinary algebra, you don't start with something you are trying to prove by assuming it is true. Then you would get arguments like this: To prove 3 = 5 start with 3 = 5. Multiply both sides by 0 and get 0 = 0. That doesn't prove 3 = 5.

I was told that you can't do that because there's no way to reverse that step, i.e. it is impossible to go from 0=0 to 3=5.

LCKurtz said:
In your current problem you could start with the left side and use your rules on it. You already have ##A - (A\cap B) = A \cap (A\cap B)^c##. Just keep going with it using the rules until you get an expression identical to the right side of the equation.

Alright. The textbook already has the answer since it's an example but I just figured I'd try to solve it like I do algebraic proofs. Is there a specific property that regular algebra has that set algebra is missing?

When you work backwards like that, you need reversible steps alright. In my argument:

## A = B##, ##A\cup U = B\cup U##, ##U = U##, you can't reverse the steps because given that ##A\cup U = B\cup U## that doesn't imply ##A=B##. Your argument has a similar error.

I see. Are there any operations that I can apply in set algebra that are reversible?

To add another point to this discussion: In algebra if you have ##a+c = b+c## you can conclude ##a=b##, and if you have ##ac = bc## you can conclude ##a=b## as long as ##c \ne 0##. You don't have similar equations for sets. That is, if you have ##A\cup C = B\cup C## you can't conclude ##A=B## and if you have ##A\cap C = B\cap C## you can't conclude ##A=B##. So cupping or capping something with both sides of an equation isn't going to give you a reversible argument.

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## 1. Why is my proof of this set identity incorrect?

There can be multiple reasons why your proof of a set identity is incorrect. Some common reasons include incorrect application of set operations, missing steps, or incorrect assumptions.

## 2. How can I check if my proof of a set identity is correct?

One way to check the correctness of your proof is to work backwards and see if the starting set and the final set obtained after applying the set operations are equal. You can also ask a colleague or mentor to review your proof for any errors.

## 3. Can I use examples to prove a set identity?

Yes, using examples can be a helpful way to understand and illustrate a set identity. However, it is important to remember that examples do not constitute a proof and should not be used as the sole method of proving a set identity.

## 4. What should I do if I am stuck while trying to prove a set identity?

If you are stuck while trying to prove a set identity, you can try breaking down the identity into smaller parts and proving each part individually. You can also consult textbooks or online resources for similar proofs or ask for guidance from a mentor or colleague.

## 5. Is there a specific format for proving a set identity?

While there is no specific format for proving a set identity, it is important to clearly state and justify each step of your proof. It is also helpful to use correct notation and symbols for set operations and elements.

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