Boolean Algebra And DeMorgan Laws -

Boolean Algebra And DeMorgan Laws -

i trying find complement of function: x(y+z!w+!vs) utilizing demorgan's laws.

i need express result sum of products.

i think complement should be: !x + (y!(z+!w)(v!s)) i'm not sure...

and if i'm uncertain how sum of product form.

thoughts?

![x(y+z!w+!vs)] = ![xy + xz!w + x!vs] distributive law = !(xy)!(xz!w)!(x!vs) de morgan's law = (!x+!y)(!x+!z+w)(!x+v+!s) de morgan's law = (!x!x+!x!z+!x!w+!y!x+!y!z+!yw)(!x+v+!s) distributive law = (!x+!y!z+!yw)(!x+v+!s) x or (x , y) = x = !x!x+!xv+!x!s+!y!z!x+!y!zv+!y!z!s+!yw!x+!ywv+!yw!s distributive law = !x+!y!zv+!y!z!s+!ywv+!yw!s x or (x , y) = x

this in sum-of-products form. taking bit further...

= !x+!y(!z+w)(v+!s) distributive law

boolean boolean-logic