1. ## Set

For two sets, P, Q, the difference P-Q is defined to be the set consisting of those elements of P that are not in Q. Prove, with the aid of de Morgan's laws, that
a) A - (B U C) = (A - B) ∩ (A - C)
b) A - (B ∩ C) = (A - B) U (A - C)

2. Here is b.
$\displaystyle \begin{array}{ll} A\backslash \left( {B \cap C} \right) & = A \cap \left( {B \cap C} \right)^c \\ & = A \cap \left( {B^c \cup C^c } \right) \\ & = \left( {A \cap B^c } \right) \cup \left( {A \cap C^c } \right) \\ & = \left( {A\backslash B} \right) \cup \left( {A\backslash C} \right) \\ \end{array}$

3. Originally Posted by Plato
Here is b.
$\displaystyle \begin{array}{ll} A\backslash \left( {B \cap C} \right) & = A \cap \left( {B \cap C} \right)^c \\ & = A \cap \left( {B^c \cup C^c } \right) \\ & = \left( {A \cap B^c } \right) \cup \left( {A \cap C^c } \right) \\ & = \left( {A\backslash B} \right) \cup \left( {A\backslash C} \right) \\ \end{array}$
What is the symbol, "\" means?

4. Originally Posted by SengNee
What is the symbol, "\" means?
it's the symbol for set subtraction. it's used the same as the minus signs you wrote in your original post

(a) is very similar, you should be able to do it now

by the way, Plato utilized DeMorgan's law in the second line of his "proof", the first line was by definition

5. a)
A - (B U C)
= A ∩ (B U C)'
= A ∩ (B' ∩ C')

I only can do until this step...

6. Originally Posted by SengNee
a)
A - (B U C)
= A ∩ (B U C)'
= A ∩ (B' ∩ C')

I only can do until this step...
just expand.

so, $\displaystyle A \cap (B' \cap C') = (A \cap B') \cap (A \cap C')$

now what?

7. Originally Posted by Jhevon
just expand.

so, $\displaystyle A \cap (B' \cap C') = (A \cap B') \cap (A \cap C')$

now what?

That means,

Distributive law not only for,

$\displaystyle \begin {array}{ll} A \cap ( B \cup C) = (A \cap B) \cup (A \cap C) \\ A \cup ( B \cap C) = (A \cup B) \cap (A \cup C) \\ \end {array}$

But also for,
$\displaystyle \begin {array}{ll} A \cap ( B \cap C) = (A \cap B) \cap (A \cap C) \\ A \cup ( B \cup C) = (A \cup B) \cup (A \cup C) \\ \end {array}$

Right?

8. Originally Posted by SengNee
That means,

Distributive law not only for,

$\displaystyle \begin {array}{ll} A \cap ( B \cup C) = (A \cap B) \cup (A \cap C) \\ A \cup ( B \cap C) = (A \cup B) \cap (A \cup C) \\ \end {array}$

But also for,
$\displaystyle \begin {array}{ll} A \cap ( B \cap C) = (A \cap B) \cap (A \cap C) \\ A \cup ( B \cup C) = (A \cup B) \cup (A \cup C) \\ \end {array}$

Right?
yes