# prooving set identities

• June 9th 2010, 11:28 PM
dunsta
prooving set identities
This is an example solution I am reading but I can't understand the 3rd transformation.

$(A-B) \cup(B-A) =(A \cup B) - (A \cap B)$
I understand changing the LHS to

$(A-B) \cup(B-A) = (A\bigcap B)\cup(B\cap A')$ < ---a (this)

using set diff representation

(The notes I am reading use this notation where sometimes the intersection operator is "big", does it have an alternate meaning when it is big, or simply a typo?)

If it is just a typo, then I needn't worry about it.

but the next step I don't understand:

(to this) b ---> $(A \cup (B \cap A')) \cap (B' \cup (B \cap A'))$
the notes claim this is using distributivity,
but I can't see how they arrive at that equation using this rule

$A \cap (B \cup C)) = (A \cap B) \cup (A \cap C)$ <-- this I understand,

but I can;t see how they went from a to b
• June 10th 2010, 02:35 AM
oldguynewstudent
Quote:

Originally Posted by dunsta
This is an example solution I am reading but I can't understand the 3rd transformation.

$(A-B) \cup(B-A) =(A \cup B) - (A \cap B)$
I understand changing the LHS to

$(A-B) \cup(B-A) = (A\bigcap B)\cup(B\cap A')$ < ---a (this)

using set diff representation

(The notes I am reading use this notation where sometimes the intersection operator is "big", does it have an alternate meaning when it is big, or simply a typo?)

If it is just a typo, then I needn't worry about it.

but the next step I don't understand:

(to this) b ---> $(A \cup (B \cap A')) \cap (B' \cup (B \cap A'))$
the notes claim this is using distributivity,
but I can't see how they arrive at that equation using this rule

$A \cap (B \cup C)) = (A \cap B) \cup (A \cap C)$ <-- this I understand,

but I can;t see how they went from a to b

In my texts $\bigcap$ is usually an intersection of sets like

$A_1 \cap A_2 \cap A_3 \cap ... \cap A_i$

I believe this is a typo and should read $(A-B) \cup(B-A) = (A\cap B')\cup(B\cap A')$ then getting from a to b is straightforward.
• June 10th 2010, 04:13 AM
dunsta
sorry, I guess the use of $\bigcap$ is a typo in the text i am studying, as the rest of the intersection operators are $\cap$ smaller size.

Actually, leaving out the ' complement symbol was my typo (still getting used to not only "sets" but "Latex" too.

$(A-B) \cup(B-A) = (A \bigcap B')\cup(B\cap A')
$

in the text.
So i still need an explanation of how to get from
$(A-B) \cup(B-A) = (A \cap B')\cup(B\cap A')$

to $(A \cup (B \cap A')) \cap (B' \cup (B \cap A'))$

as i cant see how it relates to the distributivity example I have (in the original post)
• June 10th 2010, 04:43 AM
oldguynewstudent
Quote:

Originally Posted by dunsta
sorry, I guess the use of $\bigcap$ is a typo in the text i am studying, as the rest of the intersection operators are $\cap$ smaller size.

Actually, leaving out the ' complement symbol was my typo (still getting used to not only "sets" but "Latex" too.

$(A-B) \cup(B-A) = (A \bigcap B')\cup(B\cap A')
$

in the text.
So i still need an explanation of how to get from
$(A-B) \cup(B-A) = (A \cap B')\cup(B\cap A')$

to $(A \cup (B \cap A')) \cap (B' \cup (B \cap A'))$

as i cant see how it relates to the distributivity example I have (in the original post)

This is just the distributive law of set identities on the RHS:

$A \cup (B \cap C) = (A \cap B) \cup (A \cap C)$

In the above identity replace A with $A \cap B'$ etc.