Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• February 3rd 2013, 02:53 AM
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!!!
• February 3rd 2013, 03:21 AM
ILikeSerena
Quote:

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: $(a \vee b)' = a' \wedge b'$.

Now let $a = xy$ and $b=x'y'$.
Then
$(xy \vee x'y')' = (a \vee b)' = a' \wedge b'$

Now back substitute those expressions for a and b:

$a' \wedge b' = (xy)' \wedge (x'y')'$

This is the first step.
Can you apply the 1st law of the De Morgan on (xy)' and also on (x'y')'?
• February 3rd 2013, 03:46 AM
juliie
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 :(
• February 3rd 2013, 03:48 AM
ILikeSerena
That's okay.
What did you try?
• February 3rd 2013, 03:55 AM
juliie
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.
• February 3rd 2013, 04:27 AM
ILikeSerena
Well, you can find the laws here.

Can you find the 1st law of De Morgan?
• February 3rd 2013, 04:37 AM
juliie
yes its this one (¬x)∧(¬y) = ¬(x∨y)
• February 3rd 2013, 04:42 AM
ILikeSerena
Oops. I mixed them up myself. I meant the 2nd law of De Morgan.

It says: $(\neg x) \vee (\neg y)= \neg(x \wedge y)$

or if we put it the other way around: $\neg(x \wedge y) = (\neg x) \vee (\neg y)$

The expression $(xy)'$ is actually the same as $\neg(x \wedge y)$.
It is just a different notation.

Can you now apply the 2nd law of De Morgan to (xy)'?
And also to (x'y')'?
• February 3rd 2013, 05:01 AM
juliie
so (xy)' = (x') V (y')

but (x'y')' I don't know what to do ?
• February 3rd 2013, 05:06 AM
ILikeSerena
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)?
• February 3rd 2013, 05:24 AM
juliie
¬¬x = x which is the same as (x')' I think.

and so its (x'') V (y'') = x V y ?
• February 3rd 2013, 05:42 AM
ILikeSerena
Yes! That is correct.

So $(xy \vee x'y')' = (xy)' \wedge (x'y')' = (x' \vee y') \wedge (x \vee y)$

The logical next step is to try the law of distributivity.
Care to try?
• February 3rd 2013, 05:53 AM
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?
• February 3rd 2013, 06:24 AM
ILikeSerena
Quote:

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?
• February 3rd 2013, 06:29 AM
juliie