A proof that the Halting Problem is undecidable By:- Geoffrey K. Pullum

No general procedure for bug checks succeeds.                                                          Now, I won’t just assert that, I’ll show where it leads: I will prove that although you might work till you drop, you cannot tell if computation will stop.

For imagine we have a procedure called P that for specified input permits you to see whether specified source code, with all of its faults, defines a routine that eventually halts.

You feed in your program, with suitable data, and P gets to work, and a little while later (in finite compute time) correctly infers whether infinite looping behavior occurs.

If there will be no looping, then P prints out ‘Good.’ That means work on this input will halt, as it should. But if it detects an unstoppable loop, then P reports ‘Bad!’ --- which means you’re in the soup.

Well, the truth is that P cannot possibly be, because if you wrote it and gave it to me, I could use it to set up a logical bind that would shatter your reason and scramble your mind.

Here’s the trick that I’ll use -- and it’s simple to do. I’ll define a procedure, which I will call Q, that will use P’s predictions of halting success to stir up a terrible logical mess.

For a specified program, say A, one supplies, the first step of this program called Q I devise is to find out from P what’s the right thing to say of the looping behavior of A run on A.

If P’s answer is ‘Bad!’, Q will suddenly stop. But otherwise, Q will go back to the top, and start off again, looping endlessly back, till the universe dies and turns frozen and black.

And this program called Q wouldn’t stay on the shelf; I would ask it to forecast its run on itself. When it reads its own source code, just what will it do? What’s the looping behavior of Q run on Q?

If P warns of infinite loops, Q will quit; yet P is supposed to speak truly of it! And if Q’s going to quit, then P should say ‘Good.’ Which makes Q start to loop! (P denied that it would.)

No matter how P might perform, Q will scoop it: Q uses P’s output to make P look stupid. Whatever P says, it cannot predict Q: P is right when it’s wrong, and is false when it’s true!

I’ve created a paradox, neat as can be --- and simply by using your putative P. When you posited P you stepped into a snare; Your assumption has led you right into my lair.

So where can this argument possibly go? I don’t have to tell you; I’m sure you must know. A reductio: There cannot possibly be a procedure that acts like the mythical P.

You can never find general mechanical means for predicting the acts of computing machines; it’s something that cannot be done. So we users must find our own bugs. Our computers are losers!