Consider the following input/output table (and yes, it is the same as the last question)

boolean-function
nor-operator

#1

Consider the following input/output table (and yes, it is the same as the last question):

x y f(x, y)
0 0 1
0 1 1
1 0 0
1 1 1
Create a Boolean function that is functionally complete for the set { NOR, addition, complement } and you must use the NOR operator at least once. For 5 bonus points, make it functionally complete for the set { NOR }.

Answer:

Notice f(x,y) is 0 only when x=1 and y=0, i.e. when x(comp)= 0 and y=0.
Hence, f(x,y)= x(comp) disjunction y.
Since x disjunction y = (x NOR y) NOR (x NOR y), we have,
f(x,y) = (x(comp) NOR y) NOR (x(comp) NOR y).

Again, x(comp)= x NOR x.
Substitute this in the above equation to get the functionally complete version for the set {NOR}