# Idempotency (Boolean Algebra Theorem)

1. May 27, 2013

1. The problem statement, all variables and given/known data

The idempotency theorem X∙X=X is defined for one variable, but does it apply to logic expressions with multiple variables?

Example:
ABC∙ABC = ABC

Is this valid?

2. Relevant equations

Switching algebra single-variable theorems:

X+X = X
X∙X = X

3. The attempt at a solution

My book gives single-variable and general theorems in its switching algebra section. The single-variable theorem is listed above. The general idempotency theorems are:

X+...+X = X
X∙X...∙X = X

However, it says nothing about whether or not the expression ABC∙ABC = ABC can be made.

2. May 28, 2013

### Staff: Mentor

Hi JJBladester. The dot is AND, I presume? Could you draw a truth table to test whether A.B.C = ❲A.B.C❳.❲A.B.C❳

3. May 28, 2013

Idempotency (Multiple Variables)

I use the middle dot for AND and the plus symbol for OR. Some authors use the ampersand symbol or even a period to mean AND.

I've drawn the truth table below and it appears that idempotency is valid for multi-variable terms because F3=F1&F2. Am I correct?

My professor claims that "the theorem is for single input and single output logic" but I don't know if that's correct because of my table above...

4. May 28, 2013

### milesyoung

You know that $X \land X = X$ is true, so why not just let $X = A \land B \land C$?

5. May 28, 2013

I believe what you suggested is valid. This is why I am utterly confused why my professor said that the idempotency theorem is "for single input and single output logic".

I could also re-write the original expression as:

(A∙B∙C)∙(A∙B∙C)=(A∙A)(B∙B)∙(C∙C)=A∙B∙C

Which uses the associativity theorem and then applies the idempotency theorem to single variables.

My book says "In all of the [switching algebra] theorems, it is possible to replace each variable with an arbitrary logic expression" which seems to back up what you said (let X = A∙B∙C).

6. May 28, 2013

### milesyoung

Your theorem states that $X \land X = X$ is true for any X, where X is a variable that has a truth value. $A \land B \land C$ has a truth value so it follows that $(A \land B \land C) \land (A \land B \land C) = A \land B \land C$.

7. May 29, 2013

### Staff: Mentor

It looks like you have it all figured out.

8. May 29, 2013

Yep!

Yep, thanks to you and miles. You've helped me on various things in the past. As I work my way through my degree I am always pleasantly surprised at the helpfulness of PF members.

9. May 29, 2013

### milesyoung

I'm not entirely sure what your professor meant, but you were critical and did good work in proving what you thought to be true.