Let = {a,b} and consider the language L = {we L* | the number of a's in w is equal to the number of b's in w}. The empty string, 1, belongs to L while neither of the strings a or b in S* with length one belong to L. When considering this language it is useful to consider the well-defined, total) function f : I* +Z such that, for all we ex, f(w) is the difference between the number of a's in w and the number of b's in w. Then, for all we *, 0 if w=X, f(W) = f(n) +1 ifw = pa for some string pet, if(u) - 1 if w=ub for some string u e*, and L = {we * f(w)=0}. 1. Suppose that w=0102 ...on is a string in L with positive length n such that 01 = a (so that w begins with an a), and the prefix 0102... Om does not belong to L, for every integer m such that 1 0 for every integer m such that 1 0 for every integer m such that 1