Confusion in Turings proof of the undecidability of the Entscheidungsproblem
The transliteration of the passage into modern langauge would go as follows.
We extend the language of first-order Peano arithmetic with a unary predicate $G$ (and no axioms for $G$).
For a number $n \in \mathbb{N}$, let $\overline{n}$ be the numeral
$$\underbrace{S(S(\cdots S}_{n}(0)))$$
where $S$ is the successor symbol. For example, $\overline{3} = S(S(S(0))$.
Consider a formula $\mathfrak{U}(x)$ written in this language, whose only free variable is $x$ such that, for every $n \in \mathbb{N}$:

Peano axioms prove $\mathfrak{U}(\overline{n}) \Rightarrow G(\overline{n})$, or
Peano axioms prove $\mathfrak{U}(\overline{n}) \Rightarrow \lnot G(\overline{n})$.
Peano axioms do not prove $\lnot \mathfrak{U}(\overline{n})$.

Define $\alpha : \mathbb{N} \to \{0,1\}$ by
$$\alpha(n) =
\begin{cases}
1 &amp; \text{if Peano axioms prove $\mathfrak{U}(\overline{n}) \Rightarrow G(\overline{n})$},\\
0 &amp; \text{if Peano axioms prove $\mathfrak{U}(\overline{n}) \Rightarrow \lnot G(\overline{n})$.}
\end{cases}
$$
We say that $\mathfrak{U}$ defines the sequence $\alpha$.
Intuitively, we think of $G(x)$ as stating &quot;the $x$-th digit of $\alpha$ is $1$&quot;, and of $\lnot G(x)$ as stating &quot;the $x$-th digits of $\alpha$ is $0$&quot;.
The first and second condition on $\mathfrak{U}$ ensure that $\mathfrak{U}$ always assigns $\alpha(n)$ the value $0$ or the value $1$.
The third condition ensures that $\mathfrak{U}$ never assigns both $0$ and $1$ to $\alpha(n)$ (because it follows from the first two conditions that $\lnot \mathfrak{U}(\overline{n})$ is equiprovable with $G(\overline{n}) \land \lnot G(\overline{n})$).
Example: The formula $G(x)$ defines the sequence $1, 1, 1, 1, 1, \ldots$.
Example: The formula $G(S(x))$ does not define a sequence because $G(S(0)))$ does not imply $G(0)$ and it does not imply $\lnot G(0)$. (Remember that $G$ is a primitive symbol and that we have no axioms about it.)
Example: The formula $G(0) \land \forall x \,.\, \lnot G(S(x))$ defines the sequence $1, 0, 0, 0, 0, \ldots$
Example: The formula $x \neq 0 \land \Rightarrow G(x)$ does not define the sequence because Peano axioms prove $\lnot (0 \neq 0 \land G(0)$, thus the third condition is violated. If we tried to use this formula to define a sequence, it would assign $0$ and $1$ to $\alpha(0)$ (and it would assign $1$ to all other terms of $\alpha$).
Example: The formula $G(0) \land \lnot G(S(0)) \land \forall x . G(S(S(x)))$ defines the sequence $1, 0, 1, 1, 1, 1, \ldots$
Example: The formula
$$((\exists y . x = 2 \cdot y) \Rightarrow G(x)) \land
  ((\exists y . x = S(2 \cdot y)) \Rightarrow \lnot G(x))
$$
defines the sequence $0, 1, 0, 1, 0, 1, \ldots$