Boolean Algebra question

[eyemustbecrazy]

~ before a letter indicates a bar over that letter, x-bar = ~x

If I have an x and a y going into an OR gate, then through an inverter, I
get ~x + ~y

Then a y and a z into an AND gate then through an inverter I get ~y~z

Finally an x through an inverter to get ~x

Take the ~y~z and the ~x through an OR gate I get ~x + ~y~z

Finally, put (~x + ~y) and (~x + ~y~z) through an AND gate the result is (~x
+ ~y)(~x + ~y~z)

The question I have is, can this equation be reduced further? Can FOIL be
used in this case, or is it best expressed as is?

Thanks!
Allison

 

[Gene]

Better set up a truth table.

    x   y   x+y  ~(x+y)  (~x+~y)
    0   0    0      1       1
    0   1    1      0       1
    1   0    1      0       1
    1   1    1      0       0

Do the same to #2

Oh oh, looks like you might be algebraic. Is it DeMorgan?
~(x+y) = ~x~y

#3 Ok.

The above changes the rest.

If it were correct
(~x + ~y)(~x + ~y~z) =
(~x+~x~y~z)+(~x~y+~y~z) =
~x(1+~y)+~x~y+~y~z =
~x(1+~y)+~y~z =
~x+~y~z
would have been simpler.