🙣🙜 Or How I Learned to Stop Worrying and Love this Sentence 🙞🙡
2021-11-21
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Gödel’s fixed point lemma, also known as the Diagonal lemma, is one of those lemmas. Mathematicians are weird in that we classify results in a very unusual way — and most of this is tacit but here goes: \(\renewcommand\iff\leftrightarrow\)
Lemmas are, thus, the workhorse of the working
mathematician; and the Diagonal lemma is no exception, in fact Gödel’s
first incompleteness theorem is basically a corollary (see above) of
that lemma.
What really is the diagonal lemma about? Funny question to ask before we actually even start talking about it; although we have just started talking about it, so the the argument doesn’t apply.
Mathematics is a formal system, a mathematical formula is a sequence of symbols all of which have some function. Formulas live in what we call languages, which are formal specifications of word formation rules. Much like a sentence might not be an English sentence but be a Portuguese sentence, formulas can belong to different languages.
For instance, there is a language of arithmetic; that language has terms which are thigns like \(x+y\), etc. and it has atomic formulas \(a < b\) and \(l = r\); and it has complex formulas like \[ \forall x:\exists z: x+z = 0 \]
When we interpret a formula in an environment, we assign meaning to each of the symbols and operate with them. Now, sometimes we want to say something that is self referential, such as \[ x \ \text{is an ordinal iff}\ x \ \text{is transitive and}\ \forall y\in x: y \ \text{is an ordinal} \] or “this sentence is false”. But how exactly would you go about writting a sentence that refers to itself? Gödel’s fixed point lemma comes to our rescue through indirection.
A Gödelization, or Gödel numbering is an meta-theoretical algorithm for transforming a given formula into a term of formal arithmetic. An even more technical result (Kleene, S. C. “Introduction to Metamathematics” Theorem §49.27) states that any partial recursive function are representable; that is, for any partial recursive metamathematical \(n\)-ary function \(f\) there is a \((n+1)\)-ary formula \(\phi\) such that for any numerals \(x_1\cdots x_n\), \[ \vdash f(x_1\cdots x_n) = y \iff \phi(y,x_1\cdots x_n) \] notice those \(x_1\cdots x_n\) aren’t variables, they are numerals we picked from the start, and that “\(f(...)\)” is a number we computed. The only real variable here is \(y\).
The Gödelization of a sentence is, then, the term corresponding to \(1+1+\cdots+1\) that is equal to the number we got from whatever algorithm we chose to transform formulas into numbers. \(\newcommand\quote[1]{\text“#1\text”}\) We write that term as \(\quote\phi\).
The lemma states that for any formula \(\psi(x)\) with one free variable \(x\); there is a sentence \(\phi\) such that \[ \vdash \phi \iff \psi(\quote\phi) \]
But who cares? Think about it: if we can code formulas as numbers, we can code (certain) properties of formulas as numerical sentences like \(x\leq y\), etc. So really, if \(\phi\) is something insteresting like “the formula \(x\) is not provable from the axioms of arithmetic” then we have found a sentence \(\phi\) which is logically equivalent to “\(\phi\) provable from the axioms of arithmetic”. In fact given any formulaic property, the lemma assures that there is a formula which hold exactly if it has the property.
If \(k\) is metatheoretical numeral, we can denote by \(\mathbf m\) the term \(1+\cdots+1\) \(m\) times. One may define a partial recursive function \(\sigma(\quote{\phi(x)},m)\) which takes \(\quote{\phi(x)}\) to the number \(\quote{\phi(\mathbf m)}\). Depending on what is the Gödel numbering we chose this might be straightforward or not, but it’s always painful.
This function is represented by a particular formula \(\Sigma(x,y)\); then we let \[\alpha(x) \equiv \exists y:\Sigma(y, x, x) \land \psi(y)\] and then let \[\phi\equiv\alpha(\quote\alpha)\] \[\begin{aligned} \phi &\iff \alpha(\quote{\alpha(x)})\\ &\iff \exists y: \Sigma(y, \quote{\alpha(x)}, \quote{\alpha(x)}) \land \psi(y)\\ &\iff \exists y: y = \sigma(\quote{\alpha(x)},\quote{\alpha(x)}) \land\psi(y)\\ &\iff \exists y: y = \quote{\alpha(\quote{\alpha(x)})}\land \psi(y)\\ &\quad\quad\;\text{ since substitution is defined on those entries, }\\ &\iff y = \quote{\alpha(\quote{\alpha(x)})}\land \psi(y)\\ &\iff \psi(\quote{\alpha(\quote{\alpha(x)})})\\ &\quad\quad\;\text{ now considering that $\quote\phi$ equals $\quote{\alpha(\quote{\alpha(x)})}$, }\\ &\iff \psi(\quote\phi)\\ \end{aligned}\]And so it’s been proven that \(\phi\iff\psi(\quote\phi)\) as desired.
If one is able to define a formula \(\mathfrak P\) such that \[ (\vdash \mathfrak P(\quote\psi)) \Leftrightarrow (\vdash \psi) \] then applying the lemma on the formula \(\neg\mathfrak P\) yields a sentence \(\phi\) such that \[ \vdash \phi\iff\neg\mathfrak P(\quote\phi) \] So if \(\vdash \phi\) then \(\vdash \neg\mathfrak P(\quote\phi)\) which we know to be equivalent to \(\not\vdash\phi\), which cannot be. Conversely, if \(\vdash\neg\phi\) then \(\vdash\mathfrak P(\quote\phi)\); which we know to be equivalent to \(\vdash\phi\), leading us to another contradiction.
If the theory is consistent, then we must conclude that \(\not\vdash\phi\) and \(\not\vdash\neg\phi\).