## ketz 2 years ago Simplify: f=(x'z'+xy')'

1. Hoa

is it boolean?

2. Hoa

I see it as boolean, not derivative, If it 's so, let me know. I don't want go tooo far on the wrong way, (may get ticket from 911 :( , :) )

3. perl

yes it looks boolean

4. perl

Hoa, wnat to check over a solution of mine?

5. Hoa

sure.

6. perl
7. Hoa

I'll be there now. but since we are here to help him/her. Would you please give out the answer before leaving? don't let him/her wait while we can, ok?

8. perl

yes we can simplify this

9. perl

We can use Demorgan's Theorem: ( x + y ) ' = x'y' (xy)' = x' + y'

10. perl

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

11. Hoa

one more step, friend, you have form of xx'

12. ketz

its boolean

13. perl

(x'z'+xy')' = (x'z')' (xy')' = (x'' + z'' ) (x' + y'' ) = (x + z ) (x' + y ) = xx' +zx' +xy +zy = zx' + xy + zy

14. ketz

but it seems the answer is: z(x'+y)!

15. Hoa

@ketz I am with perl. I don't get yours

16. ketz

I think the answer needs to be converted to MMF

17. Hoa

no need to get that form if the form is reducible . perl's answer is irreducible form that 's all you need

18. perl

you claim that (x'z'+xy')' = z ( x' +y ) ?

19. perl

i think thats false, plug in values for x,y,z

20. ketz

but How to convert it to Minimal Multiplicative Form (MMF)? any idea guys?

21. perl

can you double check you copied problem correctly

22. Hoa

@ketz don't combine many problems into 1. that's another process, post a new one

23. perl

Plug x=1, y=1, z = 0 into (x'z'+xy')' (1' & 0' + 1&1')' = ( 0 &1 + 1 & 0 )' = ( 0 +0 )' = 0' = 1 Now plug x=1 , y = 1 , z = 0 into z ( x' +y ) 0 ( x' +y) = 0 , because 0*x = 0 clearly they are not equal

24. perl

ketz theres a typo somewhere

25. Hoa

@perl ketz asks about the way how to get the MMF form. not the original problem. All we need is a new problem such that we get the MMF form or we can manipulate the answer to get that form. However, by that step, we make the answer more complicated than it is. We can separate it into new topic as: convert zx' + xy + zy into MMF form

26. perl

MMF is product of sums so (x'z'+xy')' = (x'z')' (xy')' = (x'' + z'' ) (x' + y'' ) = (x + z ) (x' + y )

27. Hoa

yeap

28. perl

i dont know why hes not accepting it?

29. perl

(x'z'+xy')' =/= z(x'+y)

30. perl

=/= means not equal

31. Hoa

because there is another way to get that form .

32. perl

they are not equal, i proved it

33. perl

Plug x=1, y=1, z = 0 into (x'z'+xy')' (1' & 0' + 1&1')' = ( 0 &1 + 1 & 0 )' = ( 0 +0 )' = 0' = 1 Plug x=1 , y = 1 , z = 0 into z ( x' +y ) 0 ( x' +y) = 0 , because 0*x = 0 Therefore they are not equal

34. Hoa

by multiply (x + x') into the term which doesn't have x, (y + y') into the term which doesn't have y and so on

35. Hoa

I'm with you, you are 100% right. just misunderstand together with him

36. perl

ketz writes: but it seems the answer is: z(x'+y)!

37. Hoa

ok, let's go to his new one

38. perl

wait

39. perl

so theres two forms, a product of sums, or a sum of products

40. perl

(x'z'+xy')' = (x'z')' (xy')' = (x'' + z'' ) (x' + y'' ) = (x + z ) (x' + y ) <---- Product of Sums = xx' +zx' +xy +zy = zx' + xy + zy somehow we can get xy + x'z <--- sum of products (with no terms repeating_

41. perl

This is a Boolean Calculator http://calculator.tutorvista.com/math/582/boolean-algebra-calculator.html and i plugged in ~(~x & ~z or x & ~y)

42. perl

sorry, i think wolfram has a boolean calculator, so let me find their website

43. perl
44. perl

so this is an interesting question. im going to post a new one