Originally Posted by 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?

I Meant for all 3 parts of this?

I intended the application of the distributive law to ((x' ∨ y')(x)).
And also to ((x' ∨ y')(y)).

ohhh ok so it would be

((x)(x' V y')) = (x V x') ∧ (x V y')

((y)(x' V y')) = (y V x')∧ ( y V y')

Good!

Can you find another law to apply?
One that resembles the sub expressions you have now?

is it the Absorption 1) x∧(x∨y) = x

I agree that it resembles it, but no, it does not match...

Another?

is it the of Associativity of ∨) x∨(y∨z) = (x∨y)∨z

Nope. Try complementation.

ohhh so its the x∨¬x = 1

and it will be ( y V y') = 1

but how will I do this one (y V x') ?

btw thank you soooo much!

Oops. I just noticed I missed a small mistake here.

Originally Posted by juliie
ohhh ok so it would be

((x)(x' V y')) = (x V x') ∧ (x V y')

((y)(x' V y')) = (y V x')∧ ( y V y')

It should be:
((x)(x' V y')) = (x x') V (x y')

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

Be careful!

Now, let's go back to what we have:
((xy)∨(x'y'))' = ((x x') V (x y')) ∨ ((y x') V ( y y'))

You'll need to apply the other law of complementation:
((xy)∨(x'y'))' = ((x x') V (x y')) ∨ ((y x') V ( y y')) = (0 V (x y')) ∨ ((y x') V 0)

Can you do something more with this?

can we do the Identity for ∨) x∨0 = x ?

so

(0 V (xy')) = xy'

(0V(yx')) = yx'

which means (0 V (xy')) V (0V(yx')) = xy' V yx'

and that's the proof?