Apply the Karnaugh map method to find all simple Boolean expressions for the Boolean function (w ∨ x ∨ y ∨ z)(w' ∨ x' ∨ y' ∨ z').
I started off like this... with the truth table
w x y z and everything and then.. when I reach to the table where it goes something like this.. wx,wx',w'x'and w'x and yz,yz',y'z' y'z
It sort of stuffs up from there... :\
can some one shed some light on this thanks
I would do the following in your situation: you've got an expression in conjunctive normal form (CNF). Karnaugh maps are much better suited to working in disjunctive normal form (DNF). So, the first step I would take is to convert your expression into DNF. The link I just gave you is the reverse of the process you need to do, but the reverse (converting into DNF from CNF) is entirely analogous.
Then, once you have your DNF, I would write out your Karnaugh map, and find the simplest covering expression. Make sense?