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Editorial Team
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Editorial Team
Asked: May 27, 20262026-05-27T06:15:35+00:00 2026-05-27T06:15:35+00:00

I’m going over the proof for The Halting Problem in Intro to the Theory

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I’m going over the proof for The Halting Problem in Intro to the Theory of Computation by Sipser and my main concern is about the proof below:

If TM M doesn’t know when it’s looping (it can’t accept or reject which is why a TM is Turing Recognizable for all strings), then how would could the decider H decide if M could possibly be in a loop? The same problem will carry through when TM D does its processing.

the halting problem is undecideable

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  1. Editorial Team
    Editorial Team
    2026-05-27T06:15:35+00:00Added an answer on May 27, 2026 at 6:15 am

    This is a “proof by contradiction”, a reductio ad absurdum. (Latin phrases are always good in theory classes… as long as they make sense, of course.)

    This program H is just a program with two inputs: a string representing a program for some machine, and an input. For purposes of the proof, you simply assume the program H is correct: it simply will halt and accept if M accepts with w. You don’t need to think about how it would do that; in fact, we’re about to prove it can’t, that no such program H can exist, …

    BECAUSE

    if such a program existed, we could immediately construct another program H’ that H couldn’t decide. But, by the assumption, there is no such program: H can decide everything. So, we’re forced to conclude that no program defined as we defined H is possible.

    By the way, the reductio method of proof is more controversial than you might expect, considering how often its used, especially in Computer Science. You shouldn’t be embarrassed to find it a little odd. The magic term is “non-constructive” and if you feel really ambitious, ask one of your professors about Errett Bishop’s critique of non-constructive mathematics.

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