Is there any one can show the process or prove of question(a) and question(b)? Thanks!

The 32-bit swap after the sixteenth iteration of the DES algorithm is needed to make the encryption process invertible by simply running the ciphertext back through the algorithm with thekey order reversed. This was demonstrated in the preceding prob- lem. However, it still may not be entirely clear why the 32-bit swap is needed. To demonstrate why, solve,the following exercises. First, some notation: All B TARIL) the transformation defined by the ith iteration of the encryption TD(RL) the transformation defined by the ith iteration of the decryption THCRll)-LIR, where this transformation occurs after the sixteenth iteration a. Show that the composition TD IPIP T(Ti6(Li cquivalent to the the concatenation of the bit strings A and B algorithm for 1sIs 16 algorithm for 1s1s 16 of the encryption algorithm transformation that interchanges the 32-bit halves, L15 and R5. That is, show that b. Now suppose that we did away with the final 32-bit swap in the encryption algo- rithm. Then we would want the following equality to hold: Does it