Idempotency (Boolean Algebra Theorem)

[JJBladester]

Homework Statement

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?

Homework Equations

Switching algebra single-variable theorems:

X+X = X
X∙X = X

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.

 

[NascentOxygen]

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❳

 

[JJBladester]

Idempotency (Multiple Variables)

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)

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...

 

[milesyoung]

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

 

[JJBladester]

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

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).

 

[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.

 

[NascentOxygen]

It looks like you have it all figured out.

 

[JJBladester]

Yep!

It looks like you have it all figured out.

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.

 

[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.