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)

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- Jan 7th 2008, 01:54 AM #1

- Jan 7th 2008, 03:29 AM #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}$

- Jan 7th 2008, 10:15 PM #3

- Jan 7th 2008, 10:37 PM #4

- Jan 8th 2008, 12:43 AM #5

- Jan 8th 2008, 09:14 AM #6

- Jan 8th 2008, 07:45 PM #7

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?

- Jan 8th 2008, 08:52 PM #8