a) Let n be the pumping lemma constant. Then if L is regular, PL implies that s can be decomposed into xyz, |y| > 0, |xy| ≤n, such that xy i z is in L for all i ≥0.
Since the length of xy ≤n, y consists of all b's Then xy 2 z = anbncn, where the length of of y = j. We know j > 0 so the length of the pumped string contains at as many a's as b's as c's, and is not in L. This is a Contradiction L = {w :| n a (w) = n b (w) = nc(w)}
b)
1. y consists of all a's
Pumping y will lead to a string with more than n a's -- not in L
2. y consists of all b's
Pumping y will lead to a string with more than m b's, and leave
the number of c's untouched, such that there are no longer 2n more c's than b's -- not in L
3. y consists of a's and b's
Pumping y will lead to a string with b's before a's, -- not in L