Show that this holds in every Boolean Algebra Prove that this holds in every boolean algebra:
$$ x\land y \land z' =x \space \space \text{iff}\space \space x \lor y =y \space \space and \space \space  x\land z=0 $$
My guess is to start first with the right side:
$$  x\land y \land z' =x \space \space \Rightarrow \space \space x \lor y =y \space \space and \space \space  x\land z=0 $$ 
Now to prove the first argument of "and":
$$ x \lor y = (x \land y)\lor y=y \lor(x \land y)=(y \lor x)\land(y \lor y)=1\land y=y$$
I have skipped some steps, but I guess that is it.
And I don't know how to continue, please help.

In a Boolean algebra, $x\le z'\iff x\wedge z=0$. So the proposition is equivalent to $$x\le y\wedge z' \iff x\le y \textrm{ and }x\le z'$$ which is obvious.
