Proving a Logic Equation I have two information.
$x+y = 1$ and $xy = 0$.
Now,I need to prove this equation : $xz + x'y + yz = y + z$
What I tried:

$z(x+y) + x'y = z + x'y$

Thats all
What do you think?
 A: xz+x′y+yz = xz+ xy + x′y+yz .... (adding xy=0)
      = xz + y(x + x') +yz

      = xz + y + yz  .....( x+x' always = 1)

      = z(x+y) + y ...  rearranging terms

      = z + y ........... as x + y = 1 is given.

A: If (x+y)=1, then x=1 or y=1 or both x=1 and y=1.  Now suppose also that xy=0.  Consequently, x=y=1 is not true.  Thus exclusively x=1 or y=1.  In other words one of the following two cases hold:
Case 1: x=1, and y=0.
Case 2: x=0, and y=1.
Now let's look at the first case.  This means we'll substitute x with 0, and y with 1.
Case 1: [xz+(x′y+yz)]=[1z+(1'0+0z)].  Since 1z=z, x0=0, and 0z=0 we then obtain
[1z+(1'0+0z)]=[z+(0+0)].  Since, [z+(0+0)]=z, we then obtain
[z+(0+0)]=z.  Also, (y+z)=(0+z)=z.  So, by repeated applications of the transitive property of "=" we have that [xz+(x′y+yz)]=(y+z) in this case.    
Case 2: [xz+(x′y+yz)]=[0z+(0'1+1z)].  Since 0z=0, 0'=1, 1z=z we obtain
[0z+(0'1+1z)]=[z+(11+z)].  Since 11=1 we then obtain
[z+(11+z)]=[z+(1+z)].  Since [x+(1+y)]=1 we then obtain
[z+(1+z)]=1.  Also, (y+z)=(1+z)=1.  So, by repeated applications of the transitive property of "=" in this case we have that [xz+(x′y+yz)]=(y+z).
Since one of those cases will hold, in follows that in Boolean Algebra, in general 
[xz+(x′y+yz)]=(y+z).      
