Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 25 3 Solution Created 2026-10-03 Updated 2026-10-06
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 or , where does not contain . 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 , it specifies the unique remainder of modulo . In particular the quotient and remainder variables are bounded by arithmetic terms; division is not a new language symbol.
Every finite list can be coded by some pair . Choose larger than every list entry and divisible by every integer from to . The moduli are pairwise coprime. Indeed a prime dividing both cannot divide , but then divides ; since and is divisible by that integer, this is impossible. The Chinese remainder theorem gives with for every . Since , each holds. This is Gödel beta-function sequence coding. The external choice of a divisible 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 , and . These formulas need no auxiliary variables. For function composition in recursion theory, suppose have representations with bounded arithmetic formulas . A bounded arithmetic formula for isTreat every as an auxiliary variable in the outer existential block. The logical conjunction is true for some such variables exactly when is the desired output.
Now supposeBy structural induction for primitive recursive functions, assume and are bounded arithmetic formulas representing . We code the values with . For finite witness array coding, give each component of the step witness another code pair , so that the witness at step is the remainder specified by .
With these code pairs and 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 , their empty block and logical conjunction are omitted. Expanding every occurrence of leaves only bounded quantifiers, logical conjunctions, the bounded arithmetic formulas already obtained, and the symbols .
If , take the actual finite computation values and select witnesses for the finitely many true instances of , together with witnesses for . The Gödel beta-function sequence coding provides and the finitely many . These make true. Conversely, suppose some code pairs and initial witnesses make true. The unique remainders of give values . The initial clause makes . Every step clause makes , since its coded witnesses satisfy the representing matrix for . Induction on gives , and the final remainder clause gives . This argument includes , 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 would not give the required syntactic form.
Past exam of the mathematics course of the University of Cambridge 2021 iii Paper 120 2 Solution 2026-09-28
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 and the graphs of have existential bounded definitions. Introduce variables 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 , choose larger than all its entries and divisible by . The moduli are pairwise coprime, so the Chinese remainder theorem supplies with for every .
Supposeand the induction hypothesis gives existential bounded graph formulas for and . Then exactly when there are coding values , together with a common witness bound , such that:Write each using , bound the variables representing by , and bound every witness used by the graph formulas for and by . All quantifiers checking the displayed finite recursion are then bounded; only 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 forces its th entry to be . This proves the existential bounded representation of a primitive recursive function.