# Boolean Algebra

• February 8th 2008, 04:55 AM
algebrapro18
Boolean Algebra
Simplify this Boolean Expression

A & ~((~A) & (~B)).

According to the truth table I did it should be equivalent to ~B but I can't get the boolean algebra to work out to that.

A & ~((~A) & (~B))
= A & (~~A or ~~B) Demorgan's law
= A & (A or B) Double negation
= (A & A) or (A & B) Distributive Property

And this is where I get stuck. If A&A = just A then you get into an infinate loop of distrubutive properties and never reach your goal.
• February 8th 2008, 06:22 AM
Plato
I get that the expression is equivalent A alone.
Do a truth table on the final expression: $A \vee \left( {A \wedge B} \right)$.

You might think of it as sets: $A \cup \left( {A \cap B} \right) \equiv A$.
• February 8th 2008, 06:45 AM
Soroban
Hello, algebrapro18!

Quote:

Simplify this Boolean expression: . $A \;\cap \;\sim\![(\sim\!A) \cap (\sim\!B)]$

According to the truth table I did,
it should be equivalent to $\sim B$ . . . . no, it's equivalent to ${\color{blue}A}$

$A \;\cap \;\sim\![(\sim\!A) \cap (\sim\!B)]$

. . $= \;A \;\cap \;(\sim\sim\!A \;\cup \;\sim\sim\!B) \qquad\text{DeMorgan's law}$

. . $= \;A \;\cap\;(A \cap B)\qquad\text{Double negation}$

. . $= \;(A\cap A) \cup (A \cap B)\qquad\text{Distributive Property}$ . . . . All this is correct!

And this is where I get stuck.
If $A \cap A \:=\:A$, then you get into an infinite loop
of distrubutive properties and never reach your goal. . . . . This is true

There should be (or we must establish) a theorem

. . which states: . $A \cap (A \cup B) \;=\; A \cup (A \cap B) \;=\;A$

• February 8th 2008, 07:35 AM
algebrapro18
Yes it is equal to just A alone and we can't uses sets we have to use Boolean Algebra, so there is no graphing or ven diagrams it has to be stuff like Demorgans and Distribution and stuff. My prof. was VERY clear about that.
• February 8th 2008, 07:55 AM
Plato
Here is a proof of the absorption law. [You may have different symbols to use.]
$\left( {x \wedge y} \right) \vee x = \left( {x \wedge y} \right) \vee \left( {x \wedge 1} \right)\,\,\mbox{ idenity}.$
$\left( {x \wedge y} \right) \vee x = x \wedge \left( {y \vee 1} \right) \,\,\mbox{ distributive law}.$
$\left( {x \wedge y} \right) \vee x = x \wedge 1 \,\,\mbox{ idenity law}.$
$\left( {x \wedge y} \right) \vee x = x \,\,\mbox{ idenity law}.$

That proves that Aor(A&B)=(A&B)orA=A.