Describe the error in the following "proof" that 0"1' is not a regular language. (An error must exist because 0*1 is regular.) The proof is by contradiction. Assume that 0*1" is regular. Let p be the pumping length for 0*1 given by the pumping lemma. Choose s to be the string OP1P. You know that s is a member of 0* 1*, but Example 1.73 shows that s cannot be pumped. Thus you have a contradiction. So 0*1' is not regular. 1.30