Thread: Urgent pleaseeeeeee

Use the laws of Boolean algebra to show that the identity (xy V x'y')' = x'y V xy' holds. Give all necessary steps and state explicitly which laws you are using. You may use de Morgan’s laws without deriving them yourself. I have no idea of how to even start or which of the laws to use. Please help!!!

Originally Posted by juliie

Use the laws of Boolean algebra to show that the identity (xy V x'y')' = x'y V xy' holds. Give all necessary steps and state explicitly which laws you are using. You may use de Morgan’s laws without deriving them yourself. I have no idea of how to even start or which of the laws to use. Please help!!!

Hi juliie!

The 2nd De Morgan law states: .

Now let and .
Then

Now back substitute those expressions for a and b:

This is the first step.
Can you apply the 1st law of the De Morgan on (xy)' and also on (x'y')'?

Hey thanks for that! Problem is I don't understand how to use the laws at all, I tried but its not making sense to me

That's okay.
What did you try?

Sorry but I don't understand why we apply the law to (xy)' and (x'y')' I don't know how to show that they equal each other through algebra, I did it through the table but using the laws I missed the lecture and don't have an idea of how to do it.

Well, you can find the laws here.

Can you find the 1st law of De Morgan?

yes its this one (¬x)∧(¬y) = ¬(x∨y)

Oops. I mixed them up myself. I meant the 2nd law of De Morgan.

It says:

or if we put it the other way around:


The expression is actually the same as .
It is just a different notation.

Can you now apply the 2nd law of De Morgan to (xy)'?
And also to (x'y')'?

so (xy)' = (x') V (y')

but (x'y')' I don't know what to do ?

The same thing.

Since (xy)' = (x') V (y')

(x'y')' = ((x')(y'))' = ((x')') V ((y')') = (x'') V (y'')


There is another law called double negation.
Can you find it?
And can you apply it to this last expression (twice)?

¬¬x = x which is the same as (x')' I think.

and so its (x'') V (y'') = x V y ?

Yes! That is correct.

So

The logical next step is to try the law of distributivity.
Care to try?

distributive law is x∧(y∨z) = (x∧y)∨(x∧z)

so would it be ((x' ∧ y')(x)) V ((x' ∧ y')(y))

or does that not make sense?

Originally Posted by juliie

distributive law is x∧(y∨z) = (x∧y)∨(x∧z)

so would it be ((x' ∧ y')(x)) V ((x' ∧ y')(y))

or does that not make sense?

Yep. That's basically it. Except you have replaced occurrences of ∨ by ∧, which you shouldn't have.
It should be: ((x' ∨ y')(x)) V ((x' ∨ y')(y)).


Can you repeat the distributive law in the left part, and also in the right part?

Originally Posted by ILikeSerena

Yep. That's basically it. Except you have replaced occurrences of ∨ by ∧, which you shouldn't have.
It should be: ((x' ∨ y')(x)) V ((x' ∨ y')(y)).


Can you repeat the distributive law in the left part, and also in the right part?

so does that mean I do that law for all 3 parts of (xy \vee x'y')' = (xy)' \wedge (x'y')' = (x' \vee y') \wedge (x \vee y)