# Thread: formal proof of set

1. ## formal proof of set

A and B are subsets of X
give a formal proof that $\displaystyle (A union B)^1 = A^1 intersect B^1$

I know that $\displaystyle (A union B)^1$ contains the set of elements that are in X that are not in the union of A and B

and i know that $\displaystyle A^1 intersect B^1$ contains the set of elements resulting from the intersection of the elements that are in X and not in A or B

How can i build a formal proof from this?

2. Originally Posted by murielx
A and B are subsets of X
give a formal proof that $\displaystyle (A union B)^1 = A^1 intersect B^1$

I know that $\displaystyle (A union B)^1$ contains the set of elements that are in X that are not in the union of A and B

and i know that $\displaystyle A^1 intersect B^1$ contains the set of elements resulting from the intersection of the elements that are in X and not in A or B

How can i build a formal proof from this?

What are "$\displaystyle A^1$" and "$\displaystyle B^1$". Do you mean A' and B', the complements of A and B?

To prove "set X = set Y" prove "set X is a subset of set Y" and then prove "set Y is a subset of set X"

To prove "set X is a subset of set Y" start by saying "if a is a member of X" and use whatever you know about X and Y to conclude "a is a member of Y".

To prove $\displaystyle (A\cup B)'= A'\cap B'$
(By the way $\displaystyle \cup$ is "\cup" and $\displaystyle \cap$ is "\cap".)

you need to prove that $\displaystyle (A\cup B)'\subseteq A'\cap B'$ and then prove $\displaystyle A'\cap B'\subseteq (A\cup B)'$.

To prove the first: If $\displaystyle a\in (A\cup B)'$ then a is NOT in $\displaystyle A\cup B$. That means that a is NOT in either A or B: $\displaystyle a\in A'$ and $\displaystyle a\in B'$. From that it follows that $\displaystyle a\in A'\cap B'$ and so $\displaystyle a\in (A\cup B)'$

3. so to prove the second, i can write.

if $\displaystyle a \in A' \cap B'$ then a is not in $\displaystyle A \cap B.$
that means that a is not in A and B.
$\displaystyle a \in A'$ and $\displaystyle a \in B'.$
from that it follows that $\displaystyle a \in (A \cup B)'$ and $\displaystyle a \in A' \cap B'$

does the last bit follow because of De morgans law?