Example 1: Consider L = {0m1m2m | m ≥ 1}.
Pick n of the pumping lemma. Pick z = 0n1n2n.

Break z into uvwxy, with |vwx| ≤ n and vx != ε.
Hence vwx cannot involve both 0s and 2s, since
the last 0 and the first 2 are at least n + 1 positions apart. There are two cases:

• vwx has no 2s. Then vx has only 0s and
1s. Then uwy, which would have to be in
L, has n 2s, but fewer than n 0s or 1s.
• vwx has no 0s. Analogous.
Hence L is not a CFL.

Hence vwx cannot involve both 0s and 2s, since
the last 0 and the first 2 are at least n + 1 positions apart.