1
$\begingroup$

Could someone please explain me why

$x.y+x.z+y'.z$

Is equal to

$x.y+y'.z$?

I just can't simplificate it..

$\endgroup$
1
$\begingroup$

$x.y+x.z+y'.z$

$=x.y+x.z.1+y'z$ [as,$a.1=a$]

$=x.y+x.z.(y+y')+y'z$ [as $(a+a')=1$]

$=x.y+x.z.y+x.z.y'+y'z$

$=x.y(1+z)+y'z(x+1)$

$=(x.y+y'z)$ [as $(a+1)=1$]

Hope you got your answer.

$\endgroup$
  • $\begingroup$ Yes this helps a lot thanks! So actually in xyz is xy = xyz + xyz'. I have learned about this in functions but I didn't think of using it anywhere else $\endgroup$ – Sinan Samet Jun 6 '16 at 16:18
  • $\begingroup$ Yes you can see line 2,3 how this is applied $\endgroup$ – Hailey Jun 6 '16 at 16:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.