# Boolean Algebra Equivalencies: Can These Expressions Be Simplified?

• twoski
In summary: Or, if you draw out a Karnaugh map, you can see that both xy' and x'y' are in the same group of four, allowing you to factor out the y'.In summary, the conversation involved proving the following through boolean algebra: x’y’z’ + xy’z + x’yz’ + xyz + xyz’ = x’z’ + yz’ + xz, x’y’w + x’yw + x’yzw’ = x’w + x’yz, and xy’z + x’y’z + xyz. The conversation also touched on simplifying expressions using boolean identities and laws, such as the commutative, distributive, inverse, and idempotent
twoski

## Homework Statement

Prove the following through boolean algebra

x’y’z’ + xy’z + x’yz’ + xyz + xyz’ = x’z’ + yz’ + xz

x’y’w + x’yw + x’yzw’ = x’w + x’yz

xy’z + x’y’z + xyz

## The Attempt at a Solution

Well i start by using the commutative law on the first one.

x’y’z’ + xyz + xy’z + x’yz' + xyz’ = x’z’ + yz’ + xz

At first i thought x’y’z’ + xyz might be something i can simplify, however it doesn't look like there are any boolean identities i can use on that.

With the second and third equivalency, i don't even know where to start. :/

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For the second start by factoring x'w out of the first two terms.

LCKurtz said:
For the second start by factoring x'w out of the first two terms.

Okay, so if i shuffle things around i get

y'x'w + yx'w

Could i use the idempotent law to get rid of x'w + x'w? I guess part of my confusion comes from not knowing how to use these laws when there are more variables involved.

LCKurtz said:
For the second start by factoring x'w out of the first two terms.

twoski said:
Okay, so if i shuffle things around i get

y'x'w + yx'w

Could i use the idempotent law to get rid of x'w + x'w? I guess part of my confusion comes from not knowing how to use these laws when there are more variables involved.

Apparently "shuffle things around" means use the commutative law for AND. It looks to me like you just copied the first two terms down. You haven't done what I suggested yet.

So i assume the distributive law allows us to do something like (y' + y)x'w

And of course it follows that y' + y = 1 by the inverse law, and (1)x'w can be simplified to x'w.

So now we have... x'w + x’yzw’ = x’w + x’yz

All i see is x' that can be factored...

x'( w + yzw' )

And I'm not sure if this is right, but we can simplify to:

x'( (1)yz ) = x'( yz ) = x'yz

Which means we'd also have to do the same to the RHS...

x'( w + yz )

But you can't do anything with that... So i get stuck there. Attempt 2:

x'w + x’yzw’ = x’w + x’yz

So obviously the only thing on the LHS that differs from the RHS is the w' so i want to try and get rid of it.

I'm going to guess x'w + x'w'yz can be simplified to x' + x'yz which is just x'yz... The same result.

The RHS has me confused with that w.

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twoski said:
So i assume the distributive law allows us to do something like (y' + y)x'w

And of course it follows that y' + y = 1 by the inverse law, and (1)x'w can be simplified to x'w.

So now we have... x'w + x’yzw’ = x’w + x’yz

All i see is x' that can be factored...

x'( w + yzw' )

Good to there. Write that quantity in parentheses as (w + w'(yz)) and think about what a form (a + a'b) simplifies to.

The only law i can see that applies to (w + w'(yz)) is the inverse law, which just leaves me with x'yz again... Is that the goal?

If the LHS is x'yz then the RHS needs to be similar to that...

x'w + x'yz can simplify to x'(w + yz) but there's no way for me to get rid of w.

twoski said:
x'w + x'yz can simplify to x'(w + yz) but there's no way for me to get rid of w.

How does that last expression compare to the RHS of the given problem?

If both sides have x'yz then i can just cancel them out and I'm left with x'w... Is that my simplified expression?

I've also made progress with the first question.

x’y’z’ + xy’z + x’yz’ + xyz + xyz’ = x’z’ + yz’ + xz

We factor:

x’y’z’ + xy’z + x’yz’ + xy(z + z') = x’z’ + yz’ + xz

By the inverse law:

x’y’z’ + xy’z + x’yz’ + xy = x’z’ + yz’ + xz

We factor again (after applying the commutative law):

x'z'(y + y') + xy'z + xy = x’z’ + yz’ + xz

By the inverse law:

x'z' + xy'z + xy = x’z’ + yz’ + xz

We factor again:

x'z' + x(y'z + y) = x’z’ + yz’ + xz

By the inverse law:

x'z' + xz = x’z’ + xz + yz’

So the first question leaves me with yz' if i cancel the similar expressions out...

Also, what can i do if i have the following (from a related question):

z(xy' + x'y' + xy)

I can't factor any further... Could i just totally remove x'y' + xy since they are canceled by the inverse law?

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twoski said:
1. The problem statement

x’y’w + x’yw + x’yzw’ = x’w + x’yz

twoski said:
If both sides have x'yz then i can just cancel them out and I'm left with x'w... Is that my simplified expression?

You are trying to show the left side of that top equation equals the right side. We have been working on the left side. Didn't we end up with the right side? Do you understand that it is done?

I'm going to leave the others to you.

But.. we didn't end up with the right side since the right side has 2 more variables than the left. I'm just confused as to what the next step is, i can cancel out the LHS and part of the RHS but then I'm left with x'w. I assumed proving this equivalency meant i would end up with something like x’yz = x’yz as a final result.

twoski said:
Also, what can i do if i have the following (from a related question):

z(xy' + x'y' + xy)

I can't factor any further... Could i just totally remove x'y' + xy since they are canceled by the inverse law?
To get a feel for what's inside the parentheses, you could draw up a truth table.

Otherwise, using algebra:

xy' + x'y' + xy

=(x + x')y' + xy

= y' + xy

= (x + 1)y' + xy

= xy' + y' + xy

= x(y' + y) + y'

= x + y'

NascentOxygen said:
To get a feel for what's inside the parentheses, you could draw up a truth table.

Otherwise, using algebra:

xy' + x'y' + xy

=(x + x')y' + xy

= y' + xy

= (x + 1)y' + xy

= xy' + y' + xy

= x(y' + y) + y'

= x + y'

Ah thanks, i managed to come to the same answer using a slightly different approach..

We factor:

y'(xz + x'z) + xyz

And we factor again:

y'( z(x + x') ) + xyz

Which simplifies to:

y'z + xyz

Factoring again:
z(y' + xy)

Distributive:

z( (y + y')(y' + x) )

Simplified:

z(y' + x)

Our result is:

zx + zy'

Now I'm just confused about the answers to the first 2 questions.

x'z' + xz = x’z’ + xz + yz’

x'yz = x'w + x'yz

I'm at the final step with them, i just can't for the life of me figure out how i am to show they are equivalent if the LHS clearly isn't equal to the RHS.

LCKurtz said:
How does that last expression compare to the RHS of the given problem?

twoski said:
The only law i can see that applies to (w + w'(yz)) is the inverse law, which just leaves me with x'yz again... Is that the goal?

If the LHS is x'yz then the RHS needs to be similar to that...

x'w + x'yz can simplify to x'(w + yz) but there's no way for me to get rid of w.

twoski said:
But.. we didn't end up with the right side since the right side has 2 more variables than the left. I'm just confused as to what the next step is, i can cancel out the LHS and part of the RHS but then I'm left with x'w. I assumed proving this equivalency meant i would end up with something like x’yz = x’yz as a final result.

Perhaps I have misunderstood what you have been thinking. We are working only on the left side. In post #5 about 5 lines down you had the LHS = x'(w+yzw'). In post #6 I said it was good to there and asked you to rewrite the quantity in parentheses like (w+w'(yz)). This would give LHS = x'(w+w'(yz)). Then I noted the quantity in parentheses has the form (a + a'b) and asked you to think about how that simplifies. This has nothing to do with the right hand side of the equation. I apparently misunderstood whether you have understood that. So what does the form a+a'b simplify to? Just answer that for me now.

a+a'b simplifies to a + b, which would leave us with LHS = x'w + x'yz. d'oh.

that law is sneaky, it seems pretty much the same as the inverse law.

This is all solved now, is it?

## 1. What is Boolean algebra?

Boolean algebra is a branch of mathematics and mathematical logic, named after mathematician George Boole, that deals with binary variables and logical operations. It is commonly used in computer science and electronic engineering to analyze and simplify logic circuits and expressions.

## 2. What are Boolean algebra equivalencies?

Boolean algebra equivalencies are logical equivalences, also known as laws or theorems, that help simplify and manipulate Boolean expressions. They are used to transform complex expressions into simpler ones, making them easier to understand and evaluate.

## 3. How are Boolean algebra equivalencies used?

Boolean algebra equivalencies are used to prove the validity of logical statements, simplify logical expressions, and to design and analyze digital circuits. They are also used in programming languages to create complex logical conditions and statements.

## 4. What are some common Boolean algebra equivalencies?

Some common Boolean algebra equivalencies include De Morgan's Laws, the Commutative Laws, the Associative Laws, the Distributive Laws, the Idempotent Laws, the Identity Laws, and the Null Laws. These equivalencies can be used to simplify and manipulate Boolean expressions in various ways.

There are many resources available online and in textbooks that cover Boolean algebra and its equivalencies in depth. You can also find tutorials, practice problems, and interactive tools that can help you understand and apply Boolean algebra equivalencies in various contexts.

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