Show that for any sets A and B ,
(AnB)'-(A'nB)=B'
Thanks for helping ..
Hello,
For two sets P and Q, $\displaystyle P-Q=P\cap Q'$
Thus $\displaystyle (A\cap B)'-(A'\cap B)=(A\cap B)'\cap (A'\cap B)'$
By de Moivre's formula, this is :
$\displaystyle [(A\cap B)\cup (A'\cap B)]'$
And maybe it's clear for you that $\displaystyle (A\cap B)\cup (A'\cap B)=B$ ?
If you don't see it :
- draw a diagram
- or expand, and use the fact that $\displaystyle A\cup A'=E$, where E is the universe.
Hello, thereddevils!
Show that for any sets A and B: .$\displaystyle
(A \cap B)'-(A' \cap B) \:=\:B'$
I'll signify a couple of axioms:
. . $\displaystyle (1)\;\;P \cap P' \:=\: \emptyset \qquad (2)\;\; P \cup \emptyset \:=\: P $
$\displaystyle \begin{array}{ccccc}
(A \cap B)' - (A' \cap B) & = & (A \cap B)' \cap (A' \cap B)' && \text{d{e}f. Subtraction} \\ \\
& = & (A' \cup B') \cap (A \cup B') && \text{DeMorgan} \\ \\
& = & (A' \cap A) \cup B' && \text{Distributive} \\ \\
& = & \emptyset \cup B' && \text{axiom (1)} \\ \\
& = & B' && \text{axiom (2)} \end{array}$