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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?

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  • $\begingroup$ What's the difference between $x\cdot z$ and $xz$? $\endgroup$ – Git Gud Oct 5 '13 at 12:34
  • $\begingroup$ They are the same thing $\endgroup$ – user63112 Oct 5 '13 at 12:36
  • $\begingroup$ what is $x'$ - another variable? The negation of $x$? $\endgroup$ – Carl Mummert Oct 5 '13 at 12:39
  • $\begingroup$ The negation of x $\endgroup$ – user63112 Oct 5 '13 at 12:39
  • $\begingroup$ Have you tried just writing out a truth table? $\endgroup$ – Carl Mummert Oct 5 '13 at 12:40
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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.
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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).

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