stylistic:tone
Details of second proof

The difficulty in the proof is step 1. The reader will be helped by realizing that Turing is not explaining his subtle handiwork. (In a nutshell: he is using certain equivalencies between the “existential-“ and “universal-operators” together with their equivalent expressions written with logical operators.)

Here's an example: Suppose we see before us a parking lot full of hundreds of cars. We decide to go around the entire lot looking for: “Cars with flat (bad) tires”. After an hour or so we have found two “cars with bad tires.” We can now say with certainty that “Some cars have bad tires”. Or we could say: “It’s not true that ‘All the cars have good tires’”. Or: “It is true that: ‘not all the cars have good tires”. Let us go to another lot. Here we discover that “All the cars have good tires.” We might say, “There’s not a single instance of a car having a bad tire.” Thus we see that, if we can say something about each car separately then we can say something about ALL of them collectively.

This is what Turing does:From M he creates a collection of machines { M 1, M 2, M 3, M 4,..., Mn } and about each he writes a sentence: “ X prints at least one 0” and allows only two “ truth values ”, True = blank or False =:0:. One by one he determines the truth value of the sentence for each machine and makes a string of blanks or:0:, or some combination of these. We might get something like this: “ M 1 prints a 0” = True AND “ M 2 prints a 0” = True AND “ M 3 prints a 0” = True AND “ M 4 prints a 0” = False,... AND “ Mn prints a 0” = False. He gets the string

BBB:0::0::0:...:0:... ad infinitum

if there are an infinite number of machines Mn. If on the other hand if every machine had produced a "True" then the expression on the tape would be

BBBBB....BBBB... ad infinitum

Thus Turing has converted statements about each machine considered separately into a single "statement" (string) about all of them. Given the machine (he calls it G) that created this expression, he can test it with his machine E and determine if it ever produces a 0. In our first example above we see that indeed it does, so we know that not all the M's in our sequence print 0s. But the second example shows that, since the string is blanks then every Mn in our sequence has produced a 0.

All that remains for Turing to do is create a process to create the sequence of Mn's from a single M.

Suppose M prints this pattern:

Turing creates another machine F that takes M and crunches out a sequence of Mn's that successively convert the first n 0's to “0-bar” (0):

He states, without showing details, that this machine F is truly build-able. We can see that one of a couple things could happen. F may run out of machines that have 0's, or it may have to go on ad infinitum creating machines to “cancel the zeros”.

Turing now combines machines E and F into a composite machine G. G starts with the original M, then uses F to create all the successor-machines M1, M2,..., Mn. Then G uses E to test each machine starting with M. If E detects that a machine never prints a zero, G prints:0: for that machine. If E detects that a machine does print a 0 (we assume, Turing doesn’t say) then G prints :: or just skips this entry, leaving the squares blank. We can see that a couple things can happen.

G will print no 0’s, ever, if all the Mn’s print 0’s, OR, G will print ad infinitum 0’s if all the M’s print no 0’s, OR, G will print 0’s for a while and then stop.

Now, what happens when we apply E to G itself?

If E(G) determines that G never prints a 0 then we know that all the Mn’s have printed 0’s. And this means that, because all the Mn came from M, that M itself prints 0’s ad infinitum, OR If E(G) determines that G does print a 0 then we know that not all the Mn’s print 0’s; therefore M does not print 0’s ad infinitum.

As we can apply the same process for determining if M prints 1 infinitely often. When we combine these processes, we can determine that M does, or does not, go on printing 1's and 0's ad infinitum. Thus we have a method for determining if M is circle-free. By Proof 1 this is impossible. So the first assertion that E exists, is wrong: E does not exist.