Boolean equation properties question

Question

Usually in calculus subjects multiplying two sides of an equation by a variable is "correct" and can be applied to get from an expression "X" to another expression "Y". If the first expression is correct, the second one should be too. Adding a certain number to both sides is also valid, too.

Is this same method correct in boolean algebra? Say I want to prove that this expression:

a) x + y = y

is equivalent to:

b) ~x + y = 1

I would have to prove two things: first, that if a) is true, then b) is always true. And second, that if b) is true, then a) is always true, too.

Suppose a) is correct

x + y = y -> Multiply both sides by ~y and get:

~y(x+y) = ~y*y -> Apply distribution and complementary's axiom:

~yx + ~yy = 0 -> Apply complementary's axiom again:

~y*x + 0 = 0 -> Apply DeMorgan's law:

~(y+ ~x) = 0 -> Put complementary on both sides:

y + ~x = 1

Until now, are all these steps correct and not made up fallacies?

Many thanks!

Answer 1

You want to show that the expression $x+y = y$ is true iff the expression $\lnot x + y = 1$.

One way is to make truth tables for all values of $x,y$ and check that the tables are the same (they are).

Here is an algebraic proof:

Suppose $x+y = y$, then $x+\lnot x + y = y+\lnot x$ and since $x+\lnot x = 1$ we have $1 = y+\lnot x$.

Now suppose $\lnot x + y = 1$, then $\lnot y(\lnot x + y) = \lnot y$ and hence $\lnot y\lnot x = \lnot y$. Now invert both sides and use De Morgan to get $x + y = y$.