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It can directly be checked that this sentence indeed says of itself that it has property p (and says nothing else), since it is built up in such a way that if we perform the substitution described in it, then we obtain the sentence itself, which is stated to have property p. Now, let s denote the open sentence between the quotation marks in (3), that is, let s be:

Then, clearly, the whole sentence (3) is s("s"). That is, the formalization process should consist of two main steps. In the first step, we have to find the formal version $\eta$ of s, and then the second step is obvious: the desired sentence $\lambda$ will simply be  $\eta(g(\eta))$ (where $g(\varphi)$ is the G"odel number of $\varphi$ and plays, of course, the formal counterpart of name of $\varphi$, and, for simplicity, I leave out of consideration the difference between numbers and their formal counterparts in the language).

This is the sentence we need. On the one hand, it does not use indexicals, on the other, it indeed says of itself that it has property p (and says nothing else), since it is built up in such a way that if we perform the substitution described in it, then we obtain the sentence itself, which is stated to have property p.

Now, let s denote the open sentence between the quotation marks in (3), that is, let s be:

Then, clearly, the whole sentence (3) is s("s"). In order to obtain the fix point lemma, we should translate it into the language of formal arithmetic. Clearly, the formalization process consists of two main steps. In the first step, we have to find the formal version $\eta$ of s, and then the second step is obvious: the desired sentence $\lambda$ will simply be  $\eta(g(\eta))$ (where $g(\varphi)$ is the G"odel number of $\varphi$ and plays, of course, the formal counterpart of name of $\varphi$, and, for simplicity, I leave out of consideration the difference between numbers and their formal counterparts in the language).

Then, clearly, the whole sentence (3) is s("s"). That is, the formalization process should consist of two steps. In the first step we have to find the formal version $\eta$ of s, and then the second step is obvious: the desired sentence $\lambda$ will simply be $\eta(g(\eta))$ (where $g(\varphi)$ is the G"odel number of $\varphi$ and plays, of course, the formal counterpart of name of $\varphi$, and, for simplicity, I leave out of consideration the difference between numbers and their formal counterparts in the language).

Now, that is all. That is the essence of the proof. What remains to do is simply translate, step by step, the ordinary language argument into the formal language of arithmetic. That is a completely mechanical task.

Then, clearly, the whole sentence (3) is s("s"). That is, the formalization process should consist of two main steps. In the first step, we have to find the formal version $\eta$ of s, and then the second step is obvious: the desired sentence $\lambda$ will simply be $\eta(g(\eta))$ (where $g(\varphi)$ is the G"odel number of $\varphi$ and plays, of course, the formal counterpart of name of $\varphi$, and, for simplicity, I leave out of consideration the difference between numbers and their formal counterparts in the language).

Now, that is all. That is the essence of the proof. What remains to do is simply translate the ordinary language argument into the formal language of arithmetic. That is a completely mechanical task.

$Q\vdash \lambda \longleftrightarrow \varphi(g(\lambda)).$

Let $Q$ be Robinson arithmetic (essentially Peano arithmetic without induction) and let the formula corresponding to the property p be $\varphi=\varphi(x)$. Then, obviously, the formal version of s is $\varphi(g[x(g(x)])$. In order to continue the formalization process, we should find a formula that can play the role of $\varphi(g[x(g(x))])$, that is, a formula $\eta=\eta(x)$ such that $\eta(g(\psi))$ is provably equivalent to $\varphi(g[\psi(g(\psi))])$ for every $\psi=\psi(x)$, or equivalently (denoting the inverse of $g$ by $g^{-1}$), for any $n \in N$,

$Q\vdash \lambda \longleftrightarrow \varphi(g(\lambda)),$

where $Q$ is Robinson arithmetic (essentially Peano arithmetic without induction).

Now, let the formula corresponding to the property p be $\varphi=\varphi(x)$. Then, obviously, the formal version of s is $\varphi(g[x(g(x)])$. In order to continue the formalization process, we should find a formula that can play the role of $\varphi(g[x(g(x))])$, that is, a formula $\eta=\eta(x)$ such that $\eta(g(\psi))$ is provably equivalent to $\varphi(g[\psi(g(\psi))])$ for every $\psi=\psi(x)$, or equivalently (denoting the inverse of $g$ by $g^{-1}$), for any $n \in N$,