1. ## Boolean Algebras

Write the dual of each statement.
(x + y)(x+1) = x + xy + y

As the books skips all of the in between steps, i can only postulate on what I'm supposed to do. (I'm thinking of using a supplemental book because this book isn't helpful)
Since the book also uses a example as a definition, I'm guesing that the dual of an equation is one that is fundamentally the same but in a different form.

What I did:
$\displaystyle (x+y)(x+1) = x + xy + y$
$\displaystyle x^2 + x + xy + y = x(y+1) + y$

However, the answer is $\displaystyle xy + x0 = x(x+y)y$

The professor lost me in the lecture, though I hope to meet with him later to work on some other problems.

Thanks!

2. Yes, when a book uses an example as a definition, it's frustrating. I think that "dual" means exchanging "and" with "or" and 0 with 1, as well as negating every variable. So the dual of $\displaystyle (x+y)(x + 1)$ is $\displaystyle \bar{x}\bar{y}+\bar{x}0$. I also think that the original equation has implicit universal quantifiers:

$\displaystyle \forall x\forall y\,(x+y)(x+1) = x + xy + y$

i.e., it is viewed as a univesal law, as opposed to an equation that one has to solve. When universally quantified,

$\displaystyle \bar{x}\bar{y} + \bar{x}0 = \bar{x}(\bar{x}+\bar{y})\bar{y}$

is equivalent to

$\displaystyle xy + x0 = x(x+y)y$

Of course, negating variables may not be a part of this particular definition of "dual".