Why does this Boolean absorption law work?

Question

It is said that $x \land (x \lor y) = x$ and $x \lor (x \land y) = x$ but I can't see how.

When I use distributive law on $x \land (x \lor y)$ I get $(x \land x) \lor (x \land y)$ which is the same as $x \lor (x \land y) = x$. And then applying distributive law on that I get $(x \lor x) \land (x \lor y)$ which is the same as $x \land (x \lor y)$ which is where I started!

So all I know is $x \land (x \lor y) = x \lor (x \land y)$ but how do I drop the $y$? How do I see that they both equal $x$?

Answer 1

For $y=0$ you have $x \land (x \lor y)=x \land x =x$ and $x\lor (x \land y)=x\lor 0=x$.

For $y=1$ you have $x \land (x \lor y)=x \land 1 =x$ and $x\lor (x \land y)=x\lor x=x$.

So, indeed $x \land (x \lor y)=x \land (x \lor y)=x$.

Answer 2

Figured it out:

$x \lor (x \land y)$

$(x \land 1) \lor (x \land y)$

$x \land (1 \lor y)$

$x \land 1$

$x$

And the other one:

$x \land (x \lor y)$

$(x \lor 0) \land (x \lor y)$

$x \lor (0 \land y)$

$x \lor 0$

$x$