Boolean algebra. For all x, y, and z in B, if x + y = x + z and x × y = x × z, then y = z.

Question

In the statements below, $B$ is a Boolean algebra with $\times$ and $+$ for binary operations and ($\bar{a}$)is the complement of $a$.

4.) For all $x$, $y$, and $z$ in $B$, if $x + y = x + z$ and $x \times y = x \times z$, then $y = z$.

I am supposed to prove this statement. This is Boolean algebra so you can't just subtract/divide $x$ from both sides. Here is how I started to prove this equation but I don't know where to go from here.

I multiplied both sides by the complement of x to get $(\bar{x})(x + y) = (\bar{x})(x + z)$. Which simplifies to $(\bar{x})(x)+(\bar{x})(y) = (\bar{x})(x)+(\bar{x})(z)$. Because a variable multiplied by its complement is zero, $(\bar{x})(x)$ is zero so the equation simplifies even more to $(\bar{x})(y) = (\bar{x})(z)$. And now I don't know where to go after this. Please help me!

Answer 1

Now add $x \times y = x \times z$ on both sides. We get $x^c \times y + x \times y = x^c \times z + x \times z$. Now apply the distributive law, so $( x^c + x ) \times y = ( x^c + x ) \times z$. But note that $x^c + x = 1$, so $1 y = 1z$ and we are done.

Answer 2

$$xy+\bar x(x+y)=xy+\bar xx+\bar xy=xy+0+\bar xy=(x+\bar x)y=1y=y;$$ if $xy=xz$ and $x+y=x+z$ then $$y=xy+\bar x(x+y)=xz+\bar x(x+z)=z.$$