Question

Let A = {aⁱb^jc^k | i > j > k}. Use the pumping lemma for context-free languages to show that A is not context-free.

Answer 1

Assuming the question says a^ib^jc^k (^ depicts power here) i>j>k;

Solution:

This is a proof by contradiction.

Let's assume that the language A is context free, so, pumping Lemma must hold for A.

Let s = a^(n+2)b^(n+1)c^n since i>j>k, here, i = n+2, j = n+1, k = n; It can be seen that |s| >= n, we must express s as  uvwxy such that |vwx| <= n, |vx| >= 1. Since, |vwx| <= n, uvwxy can be expressed in five ways:

1.  vwx is a^p for some p<=n, p>=1
2.  vwx is a^p b^q for some p+q<=n, p+q>=1
3.  vwx is b^p for some p<=n, p>=1
4.  vwx is b^p c^q for some p+q<=n, p+q>=1
5.  vwx is c^q for some i<=n, i>=1

It is clear that vwx cannot contain both "a" and "c" because |vwx| <= n. So, for all the cases, if it is a context free language

u v^i w x^i y, i >= 0, should be in the language.

Disproving all the cases:

Case 5: if i=2, "c" will be added to the string making count of "c" n+1, which is not in the language.

eg: uv(vwx)xy, uv(c^q)xy, thus having more than n c;

Case 2: if i=2 either no. of "a" > no. of "b" or no. of "b" > no. of "a".

Case 3: same as in case 5, no. of "b" = no. of "c", j>k condition doesn't hold.

Case 4: if i=0, uvwxy -> u(w)y -> u(bc)y -> say y is "c" and u is "a" -> abcc, j>k condition doesn't hold.

Case 1: vwx = a^p, let p=1, u(vxw)y -> u(a)y, y = bbbc, u=a, aabbbc, i>j>k condition doesn't hold.

So, the language is not context free.

It can also be proved like since the language a^ib^jc^k i<j<k is not context free so the above given language will also not be context free.