First, argue that F is not regular.

Second, find a winning strategy for R in the regular pumping game for F. (Remark: this shows the converse of Pumping Claim 1 is false.)

Consider the language F = {abick | i, j, k > 0 and if i = 1 then j = k}. Pumping Claim 1: If L is regular, then R has a winning strategy. (This is equivalent to the Chapter 1 pumping lemma, page 78.) Context-Free Pumping Game for L: 1. C chooses an integer p > 0. 2. N chooses a string s E L such that |s| > p. 3. C chooses strings u, v, x, y, z such that s = uvxyz, (vxyl 0. 4. N chooses an integer i > 0 such that uv xyz & L