A lossy Turing machine (LTM) is a two-tape, deterministic TM with an unusual kind of tapes. The machine always starts out with its input on one of its tapes. The machine can both read and write to each of its tapes. Also, let us assume that the tape alphabet is ? = {0, 1, t}. Assume further that the LTM can find the left end of the tape without any special symbols. If the inputstring is of length n, then after 2n steps, the tapes start behaving in the following lossy manner. On each step after the 2n th step, for at most one of the tapes, the symbol underneath the tape head can be erased (replaced with the blank symbol t). For example, if the Turing Machine has initial input w = w1...w10, then on steps 21, 22, 23, and so on, if the first head is on tape 1 at position k and second is at position j on tape 2 at this time then either the contents of square k on tape 1 OR the contents of square j on tape 2 will be replaced with a blank, or both tapes stay intact. This action of erasing the contents of a cell occurs after the tape head moved onto the cell, but before the tape head performs a read or write operation. Prove that any language decidable by a standard, single-tape, deterministic TM is decidable by an LTM. (Include all steps in your proof.) [Hint: the traditional approach to dealing with lossy data storage is to use some form of redundancy.]