I have to show that Y and Z are equal,

Y = (~ABC) + (A~B~C) + (A~BC) + (AB~C) = (~ABC) + (A~C) + (A~B)

But I don't know how to simplify this

Z = (A+B+C) ( A+B+~C) (A+~B+C) (~A+B+C)

How do i start?

(with ~ i mean invert)

 

I have to show that Y and Z are equal,

Y = (~ABC) + (A~B~C) + (A~BC) + (AB~C) = (~ABC) + (A~C) + (A~B)

But I don't know how to simplify this

Z = (A+B+C) ( A+B+~C) (A+~B+C) (~A+B+C)

How do i start?

(with ~ i mean invert)

Don't worry about simplifying, just put both into the same standard form and see if they match.

This might help: https://forum.allaboutcircuits.com/blog/boolean-logic-sop-and-pos-forms.583/

 

okey thanks!

 

Don't worry about simplifying, just put both into the same standard form and see if they match.

This might help: https://forum.allaboutcircuits.com/blog/boolean-logic-sop-and-pos-forms.583/

okey thanks!

but how can I do that if I the question is "Show with Boolean algebra that the two expressions are the same."

 

okey thanks!

but how can I do that if I the question is "Show with Boolean algebra that the two expressions are the same."

Consider if you are asked to show that the two expressions:

(x+3)(x-9)

and

(x-3)² - 36

are the same.

Could you not expand each to get

(x+3)(x-9) = x² + 3x - 9x - 27 = x² - 6x - 27

(x-3)² - 36 = x² - 6x + 9 - 36 = x² - 6x - 27

And since both expand to the same expression, conclude that the two original expressions are equivalent?

Also, keep in mind that once you get them both to the same intermediate expression, you can always walk the process backwards and use that to convert one expression into the other explicitly.