Gödel numbering for sequences

By an ingenious use of the Chinese remainder theorem, we can constructively define such a recursive function ${\displaystyle \beta }$ (using simple number-theoretical functions, all of which can be defined in a total recursive way) fulfilling the specifications given above. Gödel defined the ${\displaystyle \beta }$ function using the Chinese remainder theorem in his article written in 1931. This is a primitive recursive function.[6]

Thus, for all n and for any n-length sequence of natural numbers ${\displaystyle \langle a_{0},\dots a_{n-1}\rangle }$, there exists an appropriate natural number a, called the Gödel number of the sequence such that ${\displaystyle \beta (a,i)=a_{i}}$.[7]

Our specific solution will depend on a pairing function—there are several ways to implement the pairing function, so one method must be selected. Now, we can abstract from the details of the implementation of the pairing function. We need only to know its “interface”: let ${\displaystyle \pi }$, K, and L denote the pairing function and its two projection functions, respectively, satisfying specification

${\displaystyle K\left(\pi \left(x,y\right)\right)=x}$

${\displaystyle L\left(\pi \left(x,y\right)\right)=y}$

We shall not discuss and formalize the axiom for excluding alien objects here, as it is not required to understand the method.

We shall use another auxiliary function that will compute the remainder for natural numbers. Examples:

• ${\displaystyle \mathrm {rem} (5,3)=2}$
• ${\displaystyle \mathrm {rem} (7,2)=1}$

It can be proven that this function can be implemented as a recursive function.

Using the Chinese remainder theorem, we can prove that implementing ${\displaystyle \beta }$ as

${\displaystyle \beta (s,i)=\mathrm {rem} \left(K\left(s\right),\left(i+1\right)\cdot L\left(s\right)+1\right)}$

will work, according to the specification we expect ${\displaystyle \beta }$ to satisfy. We can use a more concise form by an abuse of notation (constituting a sort of pattern matching):

${\displaystyle \beta \left(\pi \left(x_{0},m\right),i\right)=\mathrm {rem} \left(x_{0},\left(i+1\right)\cdot m+1\right)}$

Let us achieve even more readability by more modularity and reuse (as these notions are used in computer science[8]): by defining ${\displaystyle \forall i<n}$ the sequence ${\displaystyle m_{i}=(i+1)\cdot m+1}$, we can write

${\displaystyle \beta \left(\pi \left(x_{0},m\right),i\right)=\mathrm {rem} \left(x_{0},m_{i}\right)}$.

We shall use this ${\displaystyle m_{i}}$ notation in the proof.

For proving the correctness of the above definition of the ${\displaystyle \beta }$ function, we shall use several lemmas. These have their own assumptions. Now we try to find out these assumptions, calibrating and tuning their strength carefully: they should not be said in an either superfluously sharp, or unsatisfactorily weak form.

Let ${\displaystyle a_{0},\dots a_{n-1}}$ be a sequence of natural numbers. Let m be chosen to satisfy

${\displaystyle \forall i\in {\overline {n}}\setminus \left\{0\right\}\left(i\mid m\right)}$

${\displaystyle \forall i<n\left(a_{i}<m_{i}\right)}$

The first assumption is meant as

${\displaystyle 1\mid m\land \dots \land n-1\mid m}$

It is needed to meet an assumption of the Chinese remainder theorem (that of being pairwise coprime). In the literature, sometimes this requirement is replaced with a stronger one, e.g. constructively built with the factorial function,[1] but the stronger premise is not required for this proof.[2]

The second assumption does not concern the Chinese remainder theorem in any way. It will have importance in proving that the specification for ${\displaystyle \beta }$ is met eventually. It ensures that an ${\displaystyle {\tilde {x}}}$ solution of the simultaneous congruence system

${\displaystyle x\equiv a_{i}{\pmod {m_{i}}}}$ for each i where ${\displaystyle 0\leq i\leq n-1}$

also satisfies

${\displaystyle a_{i}=\mathrm {rem} ({\tilde {x}},m_{i})}$.[5][9]

A stronger assumption for m requiring ${\displaystyle \forall i<n\;(a_{i}<m)}$ automatically satisfies the second assumption (if we define the notation ${\displaystyle m_{i}}$ as above).

In the section Hand-tuned assumptions, we required that

${\displaystyle \forall i\in {\overline {n}}\setminus \left\{0\right\}\left(i\mid m\right)}$. What we want to prove is that we can produce a sequence of pairwise coprime numbers in a way that will turn out to correspond to the Implementation of the β function.

In detail:

${\displaystyle \forall i<n,j<n\;\left(i\neq j\rightarrow \mathrm {coprime} \left(m_{i},m_{j}\right)\right)}$

remembering that ${\displaystyle \forall i<n}$ we defined ${\displaystyle m_{i}=(i+1)\cdot m+1}$.

The proof is by contradiction; assume the negation of the original statement:

${\displaystyle \exists i<n,j<n\;\left(i\neq j\land \lnot \mathrm {coprime} \left(m_{i},m_{j}\right)\right)}$

We know what “coprime” relation means (in a lucky way, its negation can be formulated in a concise form); thus, let us substitute in the appropriate way:

${\displaystyle \exists i<n,j<n\;\left(i\neq j\land \exists p\in \mathrm {Prime} \;\left(p\mid m_{i}\land p\mid m_{j}\right)\right)}$

Using a “more” prenex normal form (but note allowing a constraint-like notation in quantifiers):

${\displaystyle \exists i<n,j<n,p\in \mathrm {Prime} \;\left(i\neq j\land p\mid m_{i}\land p\mid m_{j}\right)}$

Because of a theorem on divisibility, ${\displaystyle p\mid m_{i}\land p\mid m_{j}}$ allows us to also say

${\displaystyle p\mid m_{i}-m_{j}}$.

Substituting the definitions of ${\displaystyle m_{k}}$-sequence notation, we get ${\displaystyle m_{i}-m_{j}=(i-j)\cdot m}$, thus (as equality axioms postulate identity to be a congruence relation[10]) we get

${\displaystyle p\mid (i-j)\cdot m}$.

Since p is a prime element (note that the irreducible element property is used), we get

${\displaystyle p\mid i-j\lor p\mid m}$.

Now we must resort to our assumption

${\displaystyle \forall i\in {\overline {n}}\setminus \left\{0\right\}\left(i\mid m\right)}$.

The assumption was chosen carefully to be as weak as possible, but strong enough to enable us to use it now.

The assumed negation of the original statement contains an appropriate existential statement using indices ${\displaystyle i<n\land j<n\land i\neq j}$; this entails ${\displaystyle i-j\in {\overline {n}}\setminus \left\{0\right\}}$, thus the mentioned assumption can be applied, so ${\displaystyle i-j\mid m}$ holds.

We can prove by several means [11] known in propositional calculus that

${\displaystyle \left(A\land \left(A\rightarrow B\right)\right)\rightarrow B}$

holds.

Since ${\displaystyle i-j\mid m}$, by the transitivity property of the divisibility relation, ${\displaystyle p\mid i-j\rightarrow p\mid m}$. Thus (as equality axioms postulate identity to be a congruence relation [10])

${\displaystyle p\mid m}$

can be proven.

The negation of original statement contained

${\displaystyle p\mid m_{i}}$

and we have just proved

${\displaystyle p\mid m}$.

Thus,

${\displaystyle p\mid m_{i}-\left(i+1\right)\cdot m}$

should also hold. But after substituting the definition of ${\displaystyle m_{i}}$,

${\displaystyle m_{i}-\left(i+1\right)\cdot m=1}$

Thus, summarizing the above three statements, by transitivity of the equality,

${\displaystyle p\mid 1}$

should also hold.

However, in the negation of the original statement p is existentially quantified and restricted to primes ${\displaystyle \exists p\in \mathrm {Prime} }$. This establishes the contradiction we wanted to reach.

By reaching contradiction with its negation, we have just proven the original statement:

${\displaystyle \forall i<n,j<n\;\left(i\neq j\rightarrow \mathrm {coprime} \left(m_{i},m_{j}\right)\right)}$

We build a system of simultaneous congruences

${\displaystyle x\equiv a_{0}{\pmod {m_{0}}}}$

${\displaystyle \vdots }$

${\displaystyle x\equiv a_{n-1}{\pmod {m_{n-1}}}}$

We can write it in a more concise way:

${\displaystyle \forall i<n\;\left(x\equiv a_{i}{\pmod {m_{i}}}\right)}$

Many statements will be said below, all beginning with "${\displaystyle \forall i<n\;\left(\dots \right)}$". To achieve a more ergonomic treatment, from now on all statements should be read as being in the scope of an ${\displaystyle \forall i<n\;\left(\dots \right)}$ quantification. Thus, ${\displaystyle \forall i<n(}$ begins here.

Let us chose a solution ${\displaystyle x_{0}}$ for the system of simultaneous congruences. At least one solution must exist, because ${\displaystyle m_{0},\dots m_{n-1}}$ are pairwise comprime as proven in the previous sections, so we can refer to the solution ensured by the Chinese remainder theorem. Thus, from now on we can regard ${\displaystyle x_{0}}$ as satisfying

${\displaystyle x_{0}\equiv a_{i}{\pmod {m_{i}}}}$,

which means (by definition of modular arithmetic) that

${\displaystyle \mathrm {rem} \left(x_{0},m_{i}\right)=\mathrm {rem} \left(a_{i},m_{i}\right)}$

Recall the second assumption, “${\displaystyle \forall i<n\;\left(a_{i}<m_{i}\right)}$”, and remember that we are now in the scope of an implicit quantification for i, so we don't repeat its quantification for each statement.

The second assumption ${\displaystyle a_{i}<m_{i}}$ implies that

${\displaystyle \mathrm {rem} \left(a_{i},m_{i}\right)=a_{i}}$.

Now by transitivity of equality we get

${\displaystyle \mathrm {rem} \left(x_{0},m_{i}\right)=a_{i}}$.

Our original goal was to prove that the definition

${\displaystyle \beta \left(\pi \left(x_{0},m\right),i\right)=\mathrm {rem} \left(x_{0},m_{i}\right)}$

is good for achieving what we declared in the specification of ${\displaystyle \beta }$: we want ${\displaystyle \beta \left(\pi \left(x_{0},m\right),i\right)=a_{i}}$ to hold.

This can be seen now by transitivity of equality, looking at the above three equations.

(The large scope of i ends here.)

We have just proven the correctness of the definition of ${\displaystyle \beta }$: its specification requiring

${\displaystyle \forall a_{0},\dots ,a_{n-1}\;\exists s\;\forall i<n\;\beta (s,i)=a_{i}}$

is met. Although proving this was most important for establishing an encoding scheme for sequences, we have to fill in some gaps yet. These are related notions similar to existence and uniqueness (although on uniqueness, “at most one” should be meant here, and the conjunction of both is delayed as a final result).

Our ultimate question is: what number should stand for the encoding of sequence ${\displaystyle \left\langle a_{0},\dots ,a_{n-1}\right\rangle }$? The specification declares only an existential quantification, not yet a functional connection. We want a constructive and algorithmic connection: a (total) recursive function that performs the encoding.

This gap can be filled in a straightforward way: we shall use minimalization, and the totality of the resulting function is ensured by everything we have proven till now (i.e. the correctness of the definition of ${\displaystyle \beta }$ by meeting its specification). In fact, the specification

${\displaystyle \forall a_{0},\dots ,a_{n-1}\;\exists s\;\forall i<n\;\beta (s,i)=a_{i}}$

plays a role here of a more general notion (“special function”[12]). The importance of this notion is that it enables us to split off the (sub)class of (total) recursive functions from the (super)class of partial recursive functions. In brief, the specification says that a function f [13] satisfying the specification

${\displaystyle f\left(a_{0},\dots ,a_{n-1},s\right)=0\leftrightarrow \forall i<n\;\left(\beta (s,i)=a_{i}\right)}$

is a special function; that is, for each fixed combination of all-but-last arguments, the function f has root in its last argument:

${\displaystyle \forall a_{0},\dots ,a_{n-1}\;\exists s\;\left(f\left(a_{0},\dots ,a_{n-1},s\right)=0\right)}$

Thus, let us choose the minimal possible number that fits well in the specification of the ${\displaystyle \beta }$ function:[5]

${\displaystyle g:\mathbb {N} ^{n}\to \mathbb {N} }$

${\displaystyle \left\langle a_{0},\dots ,a_{n-1}\right\rangle \longmapsto \mu a.\left[\forall i<n\;\left(\beta \left(a,i\right)=a_{i}\right)\right]}$.

It can be proven (using the notions of the previous section ) that g is (total) recursive.

If we use the above scheme for encoding sequences only in contexts where the length of the sequences is fixed, then no problem arises. In other words, we can use them in an analogous way as arrays are used in programming.

But sometimes we need dynamically stretching sequences, or we need to deal with sequences whose length cannot be typed in a static way. In other words, we may encode sequences in an analogous way to lists in programming.

To illustrate both cases: if we form the Gödel numbering of a Turing machine, then the each row in the matrix of the “program” can be represented with tuples, sequences of fixed length (thus, without storing the length), because the number of the columns is fixed.[14] But if we want to reason about configuration-like things (of Turing-machines), and specifically if we want to encode the significant part of the tape of a running Turing machine, then we have to represent sequences together with their length. We can mimic dynamically stretching sequences by representing sequence concatenation (or at least, augmenting a sequence with one more element) with a totally recursive function.[15]

Length can be stored simply as a surplus member:[5]

${\displaystyle g:\mathbb {N} ^{*}\to \mathbb {N} }$

${\displaystyle \left\langle a_{0},\dots ,a_{n-1},a_{n}\right\rangle \longmapsto \mu a.\left[a_{0}=n\land \forall i<n\;\left(\beta \left(a,i+1\right)=a_{i}\right)\right]}$.

The corresponding modification of the proof is straightforward, by adding a surplus

${\displaystyle x\equiv n{\pmod {m_{0}}}}$

to the system of simultaneous congruences (provided that the surplus member index is chosen to be 0). Also, the assumptions have to be modified accordingly.