# Boolean algebra / venn diagrams

• Oct 23rd 2009, 12:12 AM
djmccabie
Boolean algebra / venn diagrams
For any two subsets A and B of some set S,

\$\displaystyle (A \cup B) - (A \cap B) = (A - B) \cup (B - A)\$

where \$\displaystyle X-Y = X \cap Y' \$

Show that this is true by using a) Venn diagrams and b) algebraic manipulation using the properties of Boolean algebra.

Ok so im struggling to start on this. Part a i don't know how to start really. Part b I'm tempted to write

\$\displaystyle X - Y = (X - Y')\cup(Y' - X)\$

but doubt its right :/
• Jan 30th 2010, 09:31 AM
h2osprey
Draw two circles which intersect in a Venn diagram, label one \$\displaystyle A\$ and the other \$\displaystyle B\$. Figure out the corresponding regions represented by the each expression an it should become clear.

For b) use distributivity and De Morgan's:

\$\displaystyle (A \cup B) \cap (A \cap B)^c=(A \cup B) \cap (A^c \cup B^c)\$\$\displaystyle
= [(A \cup B) \cap A^c] \cup [(A \cup B) \cap B^c]
= (B \cap A^c) \cup (A \cap B^c)\$