The point p in the following figure is stuck- at-1. The output f will be

The Boolean expression AC + B$$\overline C $$ is equivalent to

A Boolean function 'f' of two variables x and y is defined as follows: f(0,0)=f(0,1)=f(1,1)=1;f(1,0)=0 Assuming complements of x and y are not available, a minimum cost solution for realizing F using only 2-input NOR gates and 2-input OR gates (each having unit cost) would have a total cost

If the functions W, X, Y and Z are as follows

W= R+$$\overline P Q + \overline R $$ S
X = $$X = PQ\overline R \,\overline S + \overline P \,\overline Q \,\overline R \,\overline S + P\overline Q \,\overline R \,\overline S $$
Y = $$RS + \overline {OR + P\overline Q + \overline {PQ} } $$
Z = $$R + S + \overline {PQ + \overline {PQR} + P\overline {QS} } $$