Question

Let M1 and M2 be arbitrary Turing machines. Show that the problem "L(M1)  L(M2)" is undecidable. Hint: Please note that 'For any Turing-Recognizable language accepted by a TM M, there exists an 'Unrestricted Grammar' G that can generate the language L(M)'. The problem that decides L(G) = empty is undecidable.

Answer 1

The problem "L(M1) ⊆ L(M2)" is undecidable because if it were decidable, it would contradict the known undecidability of the emptiness problem for unrestricted grammars.

Step-by-step explanation:

The student has asked to show that the problem L(M1) ⊆ L(M2) is undecidable, where M1 and M2 are arbitrary Turing machines. This is a question related to computability theory within the field of theoretical computer science. The problem in question refers to deciding whether the language recognized by Turing machine M1 is a subset of the language recognized by Turing machine M2.

One approach to proving the undecidability of this problem is by using a reduction from another undecidable problem. Recall that the emptiness problem for unrestricted grammars—which is to decide whether L(G) is empty for an unrestricted grammar G—is undecidable.

Since every Turing-recognizable language can be generated by unrestricted grammar, one can attempt to transform the undecidability of the emptiness problem into the undecidability of the subset problem for Turing machines.