title: Why L={wxw^R| w, x belongs to {a,b}^+ } is a regular language
body: <p>Using pumping lemma, we can easily prove that the language <code>L1 = {WcW^R|W ∈ {a,b}*}</code> is <strong>not a regular language</strong>. (the alphabet is {a,b,c}; W^R represents the reverse string W)</p>
  
  <p>However, If we replace character <code>c</code> with <code>"x"(x ∈ {a,b}+)</code>, say, <code>L2 = {WxW^R| x, W ∈ {a,b}^+}</code>, then L2 <strong>is a regular language</strong>.</p>
  
  <p>Could you give me some ideas?</p>