Hi everyone. I'm working on practice questions, and I'm supposed to algebraically prove this:

X'Y' + Y'Z + XZ + XY + YZ' = X'Y' + XZ + YZ'

Upon first glance, the terms that aren't in both sides are Y'Z and XY. So, is there an easy way to prove that these two terms are equal to 0, because then LS=RS? I don't know how to prove it the long way, either..

 

So, is there an easy way to prove that these two terms are equal to 0,

Terms will not be equal to zero; however, some terms will be included within combinations of other terms. So the general procedure would be to expand terms on one side, eliminate any duplicate terms, then recombine terms in a way to create the other side.

For instance, X'Y' = X'Y'(Z+Z') = X'Y'Z+X'Y'Z' and XY'Z+XY'Z = XY'Z

 

Hi everyone. I'm working on practice questions, and I'm supposed to algebraically prove this:

X'Y' + Y'Z + XZ + XY + YZ' = X'Y' + XZ + YZ'

Upon first glance, the terms that aren't in both sides are Y'Z and XY. So, is there an easy way to prove that these two terms are equal to 0, because then LS=RS? I don't know how to prove it the long way, either..

If the two sides are equal, then that means that the term XY has no effect. So see if you can determine WHY that term has no effect. Assume that XY = TRUE. That means that the left side is TRUE, regardless of any other terms on that side and regardless of the value of Z. It also means that the right side must also be TRUE, regardless of Z. Is it? If so, why? That will let you identify the terms on the left that, between them, are able to absorb XY.

 

There 6 or 8 boolean algebra rules, you apply them one by one until the equation is simplified and none of them can be applied.

 

Hi everyone. I'm working on practice questions, and I'm supposed to algebraically prove this:

X'Y' + Y'Z + XZ + XY + YZ' = X'Y' + XZ + YZ'

Upon first glance, the terms that aren't in both sides are Y'Z and XY. So, is there an easy way to prove that these two terms are equal to 0, because then LS=RS? I don't know how to prove it the long way, either..

Hi,

Quick question:
Are you supposed to prove that this is true always or that this is conditionally true?

 

Hi,

Quick question:
Are you supposed to prove that this is true always or that this is conditionally true?

What kind of conditions are you thinking of? Both of the extra terms on the left can be removed using Boolean algebra identities.

 

Hi,

I did not look into this in detail yet, but i'll have to wait until the OP gets the solution before i say too much more as i dont want to give anything away or mislead the OP. If the problem had a non conditional solution that's fine

 

What's the simplest way to prove this?

"Simple" can have different contexts.

Expansion of terms in the Boolean expression to a basic canonical form, elimination of duplicate terms, and recombination of terms is conceptually simple; equivalent to sorting through your grocery list. Just keep track of all the items.

Application of the Boolean Consensus Theorem is not as conceptually simple, but in practice involves merely identifying the consensus terms and eliminating them. For full credit, one would probably need to identify how the consensus term(s) originated.