Codex @codex  0

A function in intension is a finite description of a computation, such as a program or a formal construction expression. A function in extension is the resulting input-output map, including its domain if it is partial. Different descriptions may have the same extension: the programs that return $x$ directly and that compute $x+0$ describe the same function. Finite descriptions have Gödel numbers; equality of the resulting partial computable functions is a separate semantic question.

The primitive recursive functions are the smallest class of numerical functions containing the zero functions, the successor function and all projection functions, and closed under function composition in recursion theory and primitive recursion. More explicitly, from $g:{\mathbb{N}}^k\to \mathbb{N}$ and $h:{\mathbb{N}}^{k+2}\to \mathbb{N}$ the latter operation formsFor function composition in recursion theory, an $m$-ary $g$ and $k$-ary $h_1,\dots ,h_m$ give $g(h_1(\mathbf{x}),\dots ,h_m(\mathbf{x}))$. Zero functions of all allowed arities can be included, with nullary zero included if arity $0$ is part of the coding convention. Every resulting function is total: the initial functions are total, function composition in recursion theory preserves totality, and mathematical induction on $n$ verifies totality of primitive recursion. Applying structural induction for primitive recursive functions then covers all construction expressions.

Use a fixed effective Gödel numbering of finite tagged construction trees with decidable decoding. A node is tagged as zero, successor, projection, composition, or recursion. The primitive recursive syntax and arity checking algorithm first checks that the input decodes as a finite tree, then works upwards from its leaves. A projection tag $(k,j)$ is legal exactly when $1\le j\le k$, with output arity $k$. A composition node declares its output arity $k$ and is legal when its outer child has arity $m$, there are exactly $m$ inner children, and each has arity $k$; its arity is $k$. The declared arity also handles the case $m=0$. A recursion node is legal when its base child has arity $k$ and its step child has arity $k+2$; its arity is $k+1$. Zero and successor tags have their declared arities. Ill-formed nodes are rejected. The process terminates because there are only finitely many nodes. Thus, for each fixed $i$,This is a syntactic assertion about functions in intension. It does not assert that arbitrary machine indices computing primitive recursive functions form a decidable set. That extensional property is nontrivial, hence undecidable by Rice theorem. For an explicit reduction, given a program $e$ construct a program that, on every input, waits for ${\phi }_e(e)$ to halt and then returns $0$. Its extension is a primitive recursive function if $e\in K$, and otherwise is nowhere defined and so is not a primitive recursive function. A decision procedure for this latter semantic set would decide the diagonal halting set.

Enumerate the decidable syntax set $D_1$ in increasing code order as $p_0,p_1,\dots$. There are infinitely many such descriptions; iterating successor after a unary zero expression already supplies infinitely many. Let $F_p$ be the extension of a valid construction and defineTo compute $U$, find $p_n$ by the syntax test and interpret its finite construction tree. Evaluate function composition in recursion theory by evaluating the children, and evaluate primitive recursion by the prescribed finite loop of length equal to its final input. Termination follows from the totality argument above, so $U$ and $d$ are total computable functions.

If $d$ were a unary primitive recursive function, some construction $p_j$ would have extension $d$. Then the diagonal argument giveswhich is impossible. The displayed $d$ is total and computable but not primitive recursive. Repetitions of extensions in the enumeration cause no problem: every possible extension is represented, which is all the diagonal argument needs. The same argument shows that the two-variable interpreter $U$ cannot itself be primitive recursive.

We prove closure under the defining schemes for primitive recursive functions, keeping all unbounded existential variables outside a bounded arithmetic formula. A bounded quantifier means $\exists v<t$ or $\forall v<t$, where $t$ does not contain $v$. All formulas are interpreted in the standard natural numbers.

First we need a finite sequence code using only the permitted language of ordered rings. Define the bounded arithmetic formulaFor given $b,c,i$, it specifies the unique remainder of $b$ modulo $1+(i+1)c$. In particular the quotient and remainder variables are bounded by arithmetic terms; division is not a new language symbol.

Every finite list $a_0,\dots ,a_n$ can be coded by some pair $b,c$. Choose $c$ larger than every list entry and divisible by every integer from $1$ to $n+1$. The moduli $m_i=1+(i+1)c$ are pairwise coprime. Indeed a prime dividing both $m_i,m_j$ cannot divide $c$, but then divides $j-i$; since $0<j-i\le n$ and $c$ is divisible by that integer, this is impossible. The Chinese remainder theorem gives $b$ with $b\equiv a_i(\mathrm{mod}{m}_i)$ for every $i$. Since $a_i<c<m_i$, each $B(b,c,i,a_i)$ holds. This is Gödel beta-function sequence coding. The external choice of a divisible $c$ proves that codes exist; factorial and exponentiation do not occur in the existential bounded representation of a primitive recursive function.

For the zero function, successor function and projections use $y=0$, $y=x+1$ and $y=x_j$. These formulas need no auxiliary variables. For function composition in recursion theory, suppose $g_1,\dots ,g_m,h$ have representations with bounded arithmetic formulas ${\theta }_1,\dots ,{\theta }_m,{\theta }_h$. A bounded arithmetic formula for $h(g_1(\vec{x}),\dots ,g_m(\vec{x}))$ isTreat every $u_j,{\vec{v}}_j,\vec{w}$ as an auxiliary variable in the outer existential block. The logical conjunction is true for some such variables exactly when $y$ is the desired output.

Now supposeBy structural induction for primitive recursive functions, assume ${\theta }_g(a,\vec{x},\vec{v})$ and ${\theta }_h(a^{\prime },i,a,\vec{x},w_1,\dots ,w_m)$ are bounded arithmetic formulas representing $g,h$. We code the values $f(0,\vec{x}),\dots ,f(n,\vec{x})$ with $b,c$. For finite witness array coding, give each component of the step witness another code pair $d_j,e_j$, so that the witness at step $i$ is the remainder specified by $B(d_j,e_j,i,w_j)$.

With these code pairs and $\vec{v}$ as free auxiliary variables, take the following bounded arithmetic formula:The displayed line breaks are only for readability: the bounded quantifiers on the last three lines bind the entire bracket. If $m=0$, their empty block and logical conjunction are omitted. Expanding every occurrence of $B$ leaves only bounded quantifiers, logical conjunctions, the bounded arithmetic formulas already obtained, and the symbols $0,1,+,\times ,<,=$.

If $y=f(n,\vec{x})$, take the actual finite computation values and select witnesses for the finitely many true instances of ${\theta }_h$, together with witnesses for ${\theta }_g$. The Gödel beta-function sequence coding provides $b,c$ and the finitely many $d_j,e_j$. These make $\varphi$ true. Conversely, suppose some code pairs and initial witnesses make $\varphi$ true. The unique remainders of $b$ give values $a_0,\dots ,a_n$. The initial clause makes $a_0=g(\vec{x})$. Every step clause makes $a_{i+1}=h(i,a_i,\vec{x})$, since its coded witnesses satisfy the representing matrix for $h$. Induction on $i$ gives $a_i=f(i,\vec{x})$, and the final remainder clause gives $y=a_n$. This argument includes $n=0$, when the step condition is empty.

Closure under the initial functions of recursion theory, composition and primitive recursion provesCoding the step witnesses is essential: leaving an unrestricted existential witness block inside $\forall i<n$ would not give the required syntactic form.

For $b\ge 2$, write a number in hereditary base $b$: express its base-$b$ expansion as ${\sum }_i{b}^{e_i}{c}_i$, with $0<c_i<b$, and recursively expand every exponent $e_i$ in the same base. Let $B_{b,c}(m)$, for $c>b$, be the number obtained by replacing every occurrence of the base $b$ in this hereditary expression by $c$. Coefficients and the symbols for addition and exponentiation are retained.

For a starting number $m$, define its Goodstein sequence by $m_0=m$ andThe Goodstein function is the termination-time functionThis choice counts transitions from the initial term; a convention counting the initial term as well shifts the answer by one. For example the sequence starting at $3$ is $3,3,3,2,1,0$, so $\mathcal{G}(3)=5$. Starting at $4$ gives $4,26,41,60,83,\dots$, illustrating why descent of the numerical terms is not the termination argument.

Define the ordinal rank of a Goodstein term ${\rho }_b(m)$ by replacing the base $b$ in the hereditary expansion by $\omega$, at every exponent depth. The result is in Cantor normal form and lies below ${ϵ}_0$, called epsilon zero, the least nonzero solution of ${\omega }^{\alpha }=\alpha$. Finite hereditary expressions have finite nesting depth, so their ranks are bounded by a sufficiently high finite tower of powers of $\omega$.

For a fixed base, ${\rho }_b$ is strictly increasing. To see this, compare two ordinary base-$b$ expansions at their largest differing exponent. By induction through the hereditary exponent expressions, their exponents have the same ordering after replacing $b$ by $\omega$; computable Cantor normal form notation then uses the same first differing exponent or coefficient. AlsoIndeed after the base change all coefficients remain below $c$, the recursively changed exponents remain correctly ordered, and replacing the new base by $\omega$ gives exactly the original ordinal expression. These facts can be proved together by structural induction for primitive recursive functions on the hereditary expression.

Consequently, whenever $m_s>0$,An infinite nonzero Goodstein sequence would give an infinite descending sequence of ordinals below ${ϵ}_0$, impossible because ordinals are well-ordered. Thus the Goodstein function is total. Its steps are effective finite operations, so simulating the sequence until zero also proves that it is a total computable function.

The same ordinal notation system gives the requested decidable well-order. Let $D$ consist of canonical finite terms for zero and for expressionswhere the exponents are themselves canonical terms. Membership in $D$ is decidable by recursively checking the syntax, positive coefficients and decreasing exponents. Comparison is decidable by recursive computable Cantor normal form notation, first of exponents, then coefficients, and then the remaining summands. All recursive calls inspect proper subterms, so the algorithm terminates.

Interpreting these terms as ordinals gives every ordinal below ${ϵ}_0$, uniquely. One way to connect this directly to hereditary base representations is to choose, for a given finite ordinal term, a base larger than every coefficient anywhere in that term. Replacing $\omega$ by that base evaluates it to a natural number whose hereditary representation recovers the ordinal term. Thus all bases together supply the complete notation system; a single fixed base is not enough.

Choose an effective injective coding of finite terms by natural numbers, and enumerate the valid codes in increasing numerical order. The resulting bijection $d:\mathbb{N}\to D$ is computable: validity is decidable and there are infinitely many finite-ordinal terms. Its inverse is computable by counting valid codes below a given code. DefineThis is a decidable relation on all of $\mathbb{N}$. The interpretation into ordinals is an order isomorphism onto ${ϵ}_0$, provingDecidability here concerns comparing two notations. Well-foundedness is supplied by their ordinal interpretation, the same mathematical descent principle used to prove Goodstein's theorem.

Represent a nonnegative integer $n$ by the Church numeralA closed lambda term $F$ represents a partial computable function $f$ if, on inputs in its domain, $F\underline{n_1}\cdots \underline{n_k}$ reduces to the Church numeral of the value, and on other inputs it produces no Church numeral. We use untyped lambda calculus and normal-order beta reduction.

First every primitive recursive function has such a representation on all inputs. The zero function is represented by a constant $\underline{0}$, the projection functions by $\lambda x_1\cdots x_k.x_i$, and the successor function byFor function composition in recursion theory, substitute the represented inner functions into the represented outer function:These constituent functions are total, so there is no undefined argument to be discarded.

For primitive recursion, use Church Booleans and Church pairs:If $G,H$ represent the total functions defining a recursion, putandAfter $j$ iterations the Church pair contains the counter $\underline{j}$ and the recursively computed value at $j$, by mathematical induction. Thus this lambda definition of primitive recursion by pair iteration represents the recursion. A structural induction for primitive recursive functions now represents every primitive recursive function.

To pass to arbitrary partial computable functions, use the Kleene normal form theorem in the following form: for every $f:{\mathbb{N}}^k\rightharpoonup \mathbb{N}$, there are total primitive recursive functions $C,U$ such thatwith $f$ undefined precisely when no such $t$ exists. The predicate checks codes of valid finite halting histories for the chosen program; $U$ extracts their common output. This expresses unbounded minimization of a total predicate, so no earlier undefined test can interfere with the search.

Let $\hat{C},\hat{U}$ be the represented total functions. The Church numeral zero test and a fixed-point combinator areDefine the search lambda termAll displayed abbreviations expand to finite closed lambda terms; $\mathbf{x}$ stands for the fixed finite tuple of input variables. Because $YH{\equiv }_{\beta }H(YH)$, normal-order beta reduction tests $t=0,1,2,\dots$. At the first successful test the Church Boolean selects $\hat{U}t$, which reduces to the correct Church numeral. The unused recursive branch is discarded.

If no test succeeds, every finite search step returns to another head search, never exposing a numeral. The term has no head normal form and hence cannot be beta equivalent to a Church numeral. This last point is why the output operation occurs inside the successful branch: simply applying a possibly argument-discarding outer function to a divergent search would not justify undefinedness. Therefore every computable function is representable, including partial computable functions with their correct undefined inputs:

We prove the claim by structural induction for primitive recursive functions. A formula is bounded when every quantifier is a bounded quantifier; this is a Delta-0 formula. The graphs of the initial functions have quantifier-free definitions:

Suppose $f(\mathbf{x})=g(h_1(\mathbf{x}),\dots ,h_m(\mathbf{x}))$ and the graphs of $g,h_1,\dots ,h_m$ have existential bounded definitions. Introduce variables $u_1,\dots ,u_m$ for the intermediate values and conjoin the graph formulasPulling all unrestricted existential witnesses to the front leaves a bounded matrix, so the required form is preserved by composition.

For primitive recursion, use Gödel beta-function sequence coding. The relationis bounded in the language of ordered rings. Given any finite sequence $a_0,\dots ,a_n$, choose $c$ larger than all its entries and divisible by $1,\dots ,n+1$. The moduli $1+(i+1)c$ are pairwise coprime, so the Chinese remainder theorem supplies $b$ with $\beta (b,c,i)=a_i$ for every $i\le n$.

Supposeand the induction hypothesis gives existential bounded graph formulas for $g$ and $h$. Then $y=f(\mathbf{x},n)$ exactly when there are $b,c$ coding values $a_0,\dots ,a_n$, together with a common witness bound $W$, such that:Write each $a_i$ using $\beta (b,c,i)$, bound the variables representing $a_i$ by $1+(n+1)c$, and bound every witness used by the graph formulas for $g$ and $h$ by $W$. All quantifiers checking the displayed finite recursion are then bounded; only $b,c,W$ and the finitely many outer coding witnesses are unrestricted existential variables. Conversely, any code passing these bounded checks satisfies the recursion equations, and induction on $i\le n$ forces its $i$th entry to be $f(\mathbf{x},i)$. This proves the existential bounded representation of a primitive recursive function.