Examples of the use of the pumping lemma for cfl's Example 1: Let L = {0

a. v contains both 0's and 1's. Then y contains only 1's. If i = 0, w₀ = uvⁱxyⁱz = uxz = 0^k1^j2^m and, since k, j < n, the resulting string is not in L.

b. y contains both 1's and 2's. Then v contains only 1's. If i = 0, w₀ =uvⁱxyⁱxz = uxz = 0^m1^k2^j and, since k, j < m, the resulting string is not in L.
Case 2: v and y each contain only one alphabet symbol. Without loss of generality assume that y  . The cases where v is not the empty string are analogous. In all cases, the string that results from letting i = 0 is not in L.

a. v 0*, y  0⁺. Then, let i = 0 so that w₀ = uxz = 0^k1^m2^{m with k < m.}

b. v 0*, y  1⁺. Then, let i = 0 so that w₀ = uxz = 0^k1^j2<sup>m with j < m.

c. v 1*, y </sup> 1⁺. Then, let i = 0 so that w₀ = uxz = 0^m1^k2^{m with k < m.}

d. v 1*, y  2⁺. Then, let i = 0 so that w₀ = uxz = 0^m1^k2^{j with j < m.}

e. v 2*, y  2⁺. Then, let i = 0 so that w₀ = uxz = 0^m1^m2^{k with k < m.
Since we have shown that no matter how w is broken up into uvxyz, there is a value of i such that} uvⁱxyⁱz is not in L, the initial assumption that L was context-free is false.
Example 2: L = {aⁿb^jc^k | k >n, k > j}
Assume L is a CFL and let m be the pumping lemma integer. Choose w = a^mb^mc^m+1. Clearly, w  L. Then w can be decomposed as w = uvxyz such that |vxy|  m, |vy|  1, and uvⁱxyⁱz is in L for every i  0. To arrive at a contradiction, consider the following cases.
Case 1: v and y contain the same symbol

a. both contain a. Let i = 2. Then w₂ = uv²xy²z has the form a^m+jb^mc^m+1. Since j ≥ 1, w₂  L

b. both contain b. Let i = 2. Then w₂ = uv²xy²z has the form a^mb^m+jc^m+1. Since j ≥ 1, w₂  L

c. both contain c. Let i = 0. Then w₀ = uxz has the form a^mb^mc^m+1-j. Since j ≥ 1, w₀  L

a. v  a*, y  b⁺. Let |y| = j and let i = 2. w₂ = a^mb^m+jc^m+1. Since j ≥ 1, w₂  L

b. v  b*, y  c⁺. Let |y| = j and let i = 0 w₀ = uxz = a^mb^mc^m+1-j. Since j ≥ 1, w₀  L
Case 3: v or y contains two different symbols

a. v contains both a's and b's. Then y contains only b's. If i = 2, uv²xy²z = a^mb^jaⁱb^mc^m+1 and, since i, j > 1, m+1 ≤ m+j so w₂ L.