The proof of the undecidability of $A_{TM}$ in Michael Sipser's textbook* contains the definition of a Turing Machine, which accepts the encoding of a TM, if this TM doesn't accept its own encoding, and rejects it, if it does. If this TM is run on its own encoding, there is problem: it should accept if it doesn't accept and vice versa.

My problem with this proof is that it strongly resembles Russel's paradox. This paradox arises if we define a set, which contains all sets that are not members of themselves. If we ask whether this set contains itself, there is problem: it should contain itself if it doesn't contain itself and vice versa.

Russels's paradox has been eliminated from axiomatic set theory: in ZFC, it follows from the axioms that such a property doesn't define a valid set. Interestingly enough, in the theory of computation, a similar property defines a valid TM.

That's why I don't like the proof in Sipser's book. I'd like to emphasize that I know that this proof is perfectly valid, but I'd like to know if there is another proof which follows a different chain of thought, and doesn't define such a TM.

*Sipser, M.: Introduction to the Theory of Computation (2nd ed.), 2006, page 179. On this page, Sipser uses the term halting problem for the language $A_{TM}$. The proper name for this language is acceptance problem, see the footnote on page 188.

  • $\begingroup$ What precisely do you mean by undecidable? Usually this means "no Turing machine exists solving the problem", so I suspect that alternate proofs would be only trivially different. $\endgroup$ Aug 15, 2013 at 21:20
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    $\begingroup$ You don't like the proof because it resembles a paradox? Is there anything specific you don't like? Or is it maybe the case that you don't accept proofs by contradiction? I had some troubles reading your first paragraph, and that might be because you might not fully understand the proof. We assume the existence of $A_{TM}$, and then we reach a contradiction by running a different machine, $M$ which has as a subprocedure $A_{TM}$. $\endgroup$
    – Pål GD
    Aug 15, 2013 at 22:52
  • $\begingroup$ @PålGD I do understand the proof, and I do accept proofs by contradiction. And yes, the motive behind my question is just a feeling about it. It's like some people don't like computer-assisted proofs. But the subjective motive doesn't invalidate the question: Is there an alternative proof? Do you know one? $\endgroup$
    – kol
    Aug 16, 2013 at 6:18

2 Answers 2


So your problem is that we construct a Turing machine that works on Turing machines. In that case, we show that the particular Turing machine we imagine can not exist, and that is indeed the case for most interesting such machines.

However, it is clear that for example the problem

Given a Turing machine (encoding), decide whether the machine has a final state.

is decidable. Similar syntactic properties are also decidable.

Therefore, there are Turing machines that work on Turing machines, and therefore themselves. How is that possible?

One (intuitive, certainly not rigorous) argument is that such machines work on encodings of Turing machines that are available independently of the machines "themselves" (by a recursive enumerator). In particular, in order to answer a question about itself a given machine will not have to encode itself, it will rather work on its own encoding provided from the outside without ever knowing it does.

In particular, there is no "infinite wrapping": a Turing machine encoding (our input) is a finite string, even if it defines an infinite set. Note how this is different from self-referential sets (easier to see for $\{A \mid A \in A\}$): if a set is infinite, it can't "refer" to itself in a finite way.

That said, any infinite set (of Turing machines) that is decidable has a finite representation as Turing machine (by virtue of its decider) and vice versa. That is, non-computability is equivalent to a set (of Turing machine encodings) not having a Turing machine (encoding), so I think there is a vague parallel to Russel's paradox after all.


There is no paradox. We have made an assumption, namely that the halting problem is solved by some Turing machine, and using it reached a paradox. Hence our assumption must be wrong. In the same way, Russell's paradox is a proof that the collection of all sets not containing themselves is a proper class.

In both cases, the paradox is solved by concluding that some object cannot exist. In the case of Russell's paradox, we can give a description of the object - properties it satisfies - but it doesn't exist as a set. In the same way, the halting problem is definable, but there is no Turing machine that accepts it.

In recursion theory we can breathe life into it by making it an oracle, and consider the implications. Similarly, in set theory we can consider relativized versions of Russell's set (relativized to some set "sub-universe"), though due to the axiom of regularity the result isn't interesting.


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