Recall the Halting Problem: LHALT = {((M), x) | M is a program that halts on input x} Consider a modified version of the problem, the e-Halting Problem: Le-HALT = {(M) | M is a program that halts on the empty string } That is, we need to determine if a given program M halts on the empty string. Note, that unlike the standard Halting Problem we are only concerned with the behaviour of M on the empty string only. From that perspective, L-HALT to be an "easier" problem. Nonetheless, as it turns out, Le-HALT is still undecidable! Show that Le-HALT is undecidable by showing that LHALT ST LE-HALT. That is, assume that we have a decider E for Le-HALT and show how to use it to decide LHALT