we have a basis and the equation is true for all y.
15. (b) If 2 x + y − 3 = 0, then y = 3 − 2 x . The only values of x that give a
positive y are 0 and 1, so the domain of μ is {0,1}, giving a minimum
value of y = 1. Therefore,
Section 13.2
1. (b) Use C T = {a,b,c}, C N = {x}, and A = {x}. The nonterminal x is used as
a boundary between the left and right sides of the target string and the
two w's are built simultaneously by
At the end, the x is removed by
3. At every step, the only possible identification of V 1 is with the entire derived
string. This results in a doubling of the string and
5. A solution is
For example
and so on.
Section 13.3
1.
15. (b) If 2 x + y − 3 = 0, then y = 3 − 2 x . The only values of x that give a
positive y are 0 and 1, so the domain of μ is {0,1}, giving a minimum
value of y = 1. Therefore,
Section 13.2
1. (b) Use C T = {a,b,c}, C N = {x}, and A = {x}. The nonterminal x is used as
a boundary between the left and right sides of the target string and the
two w's are built simultaneously by
At the end, the x is removed by
3. At every step, the only possible identification of V 1 is with the entire derived
string. This results in a doubling of the string and
5. A solution is
For example
and so on.
Section 13.3
1.