Sets: Prove (a) (A − AB) ∪ B = A ∪ B; (b) (A ∪ B) − AB = AB^c ∪ A^cB

sydbernard

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Sets: Prove (a) (A − AB) ∪ B = A ∪ B; (b) (A ∪ B) − AB = AB^c ∪ A^cB

Let A and B be two events. Prove the following relations by the element wise method.
(a) (A − AB) ∪ B = A ∪ B.
(b) (A ∪ B) − AB = AB^c ∪ A^cB.
 
Let A and B be two events. Prove the following relations by the element wise method.
(a) (A − AB) ∪ B = A ∪ B.
(b) (A ∪ B) − AB = AB^c ∪ A^cB.
Please share your work with us ...even if you know it is wrong.

If you are stuck at the beginning tell us and we'll start with the definitions.

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Let A and B be two events. Prove the following relations by the element wise method.
(a) (A − AB) ∪ B = A ∪ B.
(b) (A ∪ B) − AB = AB^c ∪ A^cB.

To make (b) a little easier to read, you can use the superscript format: (A ∪ B) − AB = ABc ∪ AcB .

If you aren't familiar with the "element-wise method", you should start by supposing that some element x is in (A ∪ B) − AB [that is, it is in either A or B, but not in both A and B], and show that it will necessarily be in ABc ∪ AcB; and then suppose that x is in ABc ∪ AcB and show that it must be in (A ∪ B) − AB.

Please show what you can do, and tell us why you are stuck, so we can help you past that.
 
So I know that you have to prove that each is a subset of one another so for a I have

let x e AUB then x e A or x e B. If x e B then x is not an element of A. and then would have to do the same where x e A but not B, but I am confused what to do from there.
 
So I know that you have to prove that each is a subset of one another so for a I have

let x e AUB then x e A or x e B. If x e B then x is not an element of A. and then would have to do the same where x e A but not B, but I am confused what to do from there.

You're right that if x ∈ (A ∪ B) − AB, then either x ∈ B but not in A, or x ∈ A but not in B. (This doesn't follow merely from x ∈ A ∪ B, but from the entire left-hand side.)

What to do next? Look at what you have to prove! what does it mean that x ∈ ABc ∪ AcB? You are practically finished with this half of the proof.
 
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