A catalogue of proofs that various sets, functions and relations are primitive recursive 26 February 2012 at 20:57

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1 A catalogue of proofs that various sets, functions and relations are primitive recursive 26 February 2012 at 20:57 Public In this note (which is mainly intended as a 'memo' for myself) I am interested in cataloguing numerous proofs that some basic sets, functions and relations are primitive recursive. All such primitive recursive formulas have the property that they can be implemented as Unlimited Register Machine (URM) programs, i.e., they are Turing-computable. The idea in this note is to formulate proofs of Turing-computability mathematically without having to formulate URM programs explicitly, by first showing that some very simple formulas are Turing-computable, and then building up new Turingcomputable formulas from these using processes which are proven to yield Turing-computable functions from Turingcomputable functions. I want to catalogue as many examples of these proofs as I can here. The starting point is to note that the following very simple functions are Turing-computable: The proof that these are Turing-computable is immediate from the fact that they can each be implemented using the single-line URM programs 1 Z(1), 1 S(1), and 1 C(m, 1) respectively. One way to build up new Turing-computable functions from these is by substitution, defined as follows:

2 It is a basic theorem of computability theory that a function obtained by substitution from Turing-computable functions is itself Turing-computable. Another way to build up new Turing-computable functions from more basic ones is by primitive recursion, defined as follows for a function of one variable: Thus, to show that primitive recursion applies to a function of one variable, we need h(0) and an ability to compute h(n+1) given n and h(n). We represent this ability by defining an appropriate function of the form g(n, h(n)) = h(n+1). We can extend the definition of primitive recursion to functions of more than one variable as follows: It is again a basic theorem of computability theory that a function obtained by primitive recursion from other functions which are Turing-computable must itself be Turing-computable. A function is said to be primitive recursive if it can be obtained from basic primitive recursive functions using the operations of substitution and primitive recursion a finite number of times. It is a basic theorem of computability theory that every primitive recursive function is Turing-computable. To extend the concept of Turing-computability to sets, we use the concept of the characteristic function of a set:

3 A subset A of the set of natural numbers N is said to be a Turing-computable set if its characteristic function is a Turing-computable function, and it is said to be a primitive recursive set if its characteristic function is a primitive recursive function. Since every primitive recursive function is a Turing-computable function, it follows immediately that every primitive recursive set is also a Turing-computable set. We can do something similar for relations, using the concept of the characteristic function of a relation: A relation R is then said to be a Turing-computable relation if its characteristic function is a Turing-computable function, and it is said to be a primitive recursive relation if its characteristic function is a primitive recursive function. Using these concepts it is possible to prove that functions defined by cases are primitive recursive. The key theorems here are the following:

4 It is also possible to use these concepts to prove that bounded minimizations, otherwise known as bounded searches, are primitive recursive. The definition of bounded minimization (on a function) is as follows: The key theorem relating to functions defined by bounded minimization on a function is as follows: Similar considerations apply to bounded minimization on a relation:

5 I will begin by providing proofs here that the following basic functions, sets and relations are primitive recursive, and then continue proving other non-standard cases in this note over the next few months:

6 The constant function C a (n) = a is primitive recursive for each natural number a. Proof: By induction. First note that C(n) = 0 = zero(n) is primitive recursive, since the zero function is a basic primitive recursive function. This establishes the basis for induction. Next, assume that C a-1 (n) = a - 1 is primitive recursive. Then C a (n) = succ(c a-1 (n)) is also primitive recursive, since it is obtained by substitution using the primitive recursive functions succ and C a- 1(n). It follows by the principle of mathematical induction that C a (n) is primitive recursive for each natural number a.

7 2. For k 1, every constant function of k variables C aᵏ(n 1, n 2,..., n k ) = a is primitive recursive. Proof: The case k = 1 was shown to be primitive recursive in the previous proof. But for each k > 1 we have C aᵏ(n 1, n 2,..., n k ) = C a ¹(n 1 ) = C a ¹(U 1ᵏ(n 1, n 2,..., n k )) Thus for each k > 1 the function C aᵏ(n 1, n 2,..., n k ) = a is primitive recursive because it is defined by substitution from the primitive recursive function C a ¹ and the projection function U 1ᵏ which is a basic primitive recursive function. 3. The function add(n, m) = n + m is primitive recursive. Proof: Let h(n, m) add(n, m). Then we have: h(n, 0) = add(n, 0) = n h(n, m+1) = add(n, m+1) = add(n, m) + 1 Therefore we can write: h(n, 0) = f(n) = U 1 ¹(n) h(n, m+1) = g(n, m, h(n, m)) = succ(u 3 ³(n, m, h(n, m))) Therefore add(n, m) can be obtained by substitution and primitive recursion from the basic primitive recursive functions U 1 ¹, succ, and U 3 ³ and must therefore be primitive recursive itself. 4. The function mult(n, m) = n m is primitive recursive. Proof: Let h(n, m) mult(n, m). Then we have: h(n, 0) = mult(n, 0) = 0 h(n, m+1) = mult(n, m+1) = n + mult(n, m) Therefore we can write: h(n, 0) = f(n) = zero(n) h(n, m+1) = g(n, m, h(n, m)) = add(u 1 ³(n, m, h(n, m)), U 3 ³(n, m, h(n, m))) Therefore mult(n, m) can be obtained by substitution and primitive recursion from the basic primitive recursive functions zero, U 1 ³, and U 3 ³, and the primitive recursive function add. It follows that mult(n, m) must itself be primitive recursive. 5. The linear function f(n) = an + b, with a and b fixed natural numbers, is primitive recursive. Proof: We can write: f(n) = add(mult(c a (n), U 1 ¹(n)), C b (n)) Thus f(n) = an + b is primitive recursive, since it is obtained by substitution using the primitive recursive functions add, mult, C a, C b, and U 1 ¹. 6. The exponentiation function exp(n, m) = nᵐ is primitive recursive. Proof: Let h(n, m) exp(n, m). Then we have: h(n, 0) = exp(n, 0) = 1 h(n, m+1) = exp(n, m+1) = n nᵐ Therefore we can write: h(n, 0) = f(n) = C 1 (n) h(n, m+1) = g(n, m, h(n, m)) = mult(u 1 ³(n, m, h(n, m)), U 3 ³(n, m, h(n, m))) Thus exp(n, m) is primitive recursive, since it is obtained by substitution and primitive recursion using the primitive recursive functions C 1, mult, U 1 ³ and U 3 ³. 7. The predecessor function pred(n) = n -1 if n > 0, pred(n) = 0 if n = 0, is primitive recursive.

8 Proof: Let h(n) pred(n). Then we have: h(0) = 0 h(n + 1) = pred(n + 1) = n Writing h(n + 1) = g(n, h(n)) = U 1 ²(n, h(n)) we see that pred(n) is obtained by primitive recursion using the basic primitive recursive function U 1 ², and must therefore be primitive recursive itself. 8. The cut-off subtraction function n + m = n - m if m n, n + m = 0 if n < m, is primitive recursive. Proof: Let h(n, m) n + m. Then we have: h(n, 0) = n h(n, m + 1) = n + (m + 1) = (n + m) + 1 = pred(n + m) Therefore we can write: h(n, 0) = f(n) = U 1 ¹(n) h(n, m+1) = g(n, m, h(n, m)) = pred(u 3 ³(n, m, h(n, m))) It follows that cut-off subtraction is primitive recursive because it is obtained by substitution and primitive recursion using the primitive recursive functions U 1 ¹, pred and U 3 ³. 9. The functions max(n, m) and min(n, m) are primitive recursive. Proof: We have min(n, m) = n + (n + m) = U 1 ²(n, m) + (U 1 ²(n, m) + U 2 ²(n, m)) so min(n, m) is obtained by substitution using the cut-off subtraction function which is primitive recursive, and the basic primitive recursive functions U 1 ² and U 2 ². It must therefore be primitive recursive itself. But: max(n, m) + min(n, m) = n + m max(n, m) = n + m + min(n, m) = n + m + (n + (n + m)) = m + (n + m) Therefore max(n, m) = add(u 2 ²(n, m), (U 1 ²(n, m) + U 2 ²(n, m))) so max(n, m) is obtained by substitution using the add and cut-off subtraction functions which are primitive recursive, and the basic primitive recursive functions U 1 ² and U 2 ². It must therefore be primitive recursive itself. 10. The function adf(n, m) = n - m is primitive recursive. Proof: We can write adf(n, m) = (n + m) + (m + n) = add((u 1 ²(n, m) + U 2 ²(n, m)), (U 2 ²(n, m) + U 1 ²(n, m))) so adf(n, m) is obtained by substitution using the add and cut-off subtraction functions which are primitive recursive, and the basic primitive recursive functions U 1 ² and U 2 ². It must therefore be primitive recursive itself. 11. The signum function sg(n) = 1 if n > 0, sg(n) = 0 if n = 0, is primitive recursive. (This is the characteristic function of the set of positive integers). Proof: We have sg(n) = 1 + (1 + n) = C 1 (n) + (C 1 (n) + U 1 ¹(n)) Therefore sg(n) is primitive recursive because it is obtained by substitution using cut-off subtraction, which is primitive recursive, and the basic primitive recursive functions C 1 and U 1 ¹. Since sg(n) = 1 + sg(n) = C 1 (n) + sg(n) is obtained by substitution from the primitive recursive functions C 1, cut-off subtraction and sg, it is also a primtive recursive function. (This is the characteristic function of the set {0}).

9 12. The remainder function rem(n, m) is primitive recursive, where: rem(n, m) = the remainder when n is divided by m if m 0, rem(n, m) = 0 if m = 0. Proof: To prove this we use the fact that if a function h can be obtained from known primitive recursive functions by primitive recursion where a variable other than the last is taken as the recursion variable, the function h is primitive recursive. Thus, for example, we can show that h(n + 1, m) is primitive recursive instead of always having to show that h(n, m + 1) is primitive recursive. The former is more convenient than the latter in some cases, such as this one. Therefore let h(n, m) rem(n, m). Using the first variable as the recursion variable we clearly have h(0, m) = 0. We have h(n + 1, m) = 0 if m = 0 or if m = n + 1, and h(n + 1, m) = h(n, m) + 1 otherwise. A little thought shows we can write this as a single equation as: h(n + 1, m) = sg(m)sg(adf(add(n, 1), m))add(1, h(n, m)) Therefore we can write: h(0, m) = f(m) = zero(m) h(n, m+1) = g(n, m, h(n, m)) = mult(mult(sg(m), sg(adf(add(n, 1), m))), add(1, h(n, m))) It follows that the remainder function rem(n, m) is primitive recursive because it is obtained by substitution and primitive recursion (over the first variable) using the primitive recursive functions, zero, mult, sg, adf, add, C 1 ³, U 1 ³, U 2 ³, and U 3 ³ (the constant and projection functions are not shown explicitly in the expression to reduce clutter). 13. The quotient function quot(n, m) is primitive recursive, where: quot(n, m) = the quotient when n is divided by m if m 0, quot(n, m) = 0 if m = 0. Proof: Let h(n, m) quot(n, m). Using the first variable as the recursion variable we clearly have h(0, m) = 0. We have h(n + 1, m) = h(n, m) + 1 if rem(n + 1, m) = 0, and h(n + 1, m) = h(n, m) if rem(n + 1, m) 0. We can write this as a single equation as: h(n + 1, m) = sg(rem(n + 1, m))h(n, m) + (1 - sg(rem(n + 1, m)))(h(n, m) + 1) Therefore we can write: h(0, m) = f(m) = zero(m) h(n + 1, m) = add(mult(sg(rem(add(n, 1), m)), h(n, m)), mult((1 + sg(rem(add(n, 1), m))), add(h(n, m), 1))) It follows that the quotient function quot(n, m) is primitive recursive because it is obtained by substitution and primitive recursion (over the first variable) using the primitive recursive functions, zero, mult, sg, adf, add, cut-off subtraction C 1 ³, U 1 ³, U 2 ³, and U 3 ³ (the constant and projection functions are not shown explicitly in the expression to reduce clutter). 14. The factorial function fac(n) = n! is primitive recursive. Proof: Let h(n) fac(n). Then we have: h(0) = 1 h(n + 1) = g(n, h(n)) = (n + 1)h(n) = mult(add(n, 1), h(n)) It follows that fac(n) is primitive recursive since it is obtained by substitution and primitive recursion using the primitive recursive functions mult and add, and the (not shown explicitly) basic primitive recursive constant and projection functions. 15. The relation eq of equality between two natural numbers is primitive recursive. Proof: Let χ eq (n, m) = 1 if n = m, χ eq (n, m) = 0 if n m. Then we can write: χ eq (n, m) = 1 + sg(adf(n, m))

10 Therefore the characteristic function of the equality relation is a primitive recursive function, since it is obtained by substitution using the primitive recursive functions sg, adf and cut-off subtraction, and a constant. It follows that the equality relation is primitive recursive. 16. The relations, >, and < are primitive recursive. Proof: Let χ (n, m) = 1 if n m, χ (n, m) = 0 if n > m. Then we can write: χ (n, m) = 1 + sg(n + m) Therefore the characteristic function of the relation is a primitive recursive function, since it is obtained by substitution using the primitive recursive functions sg, cut-off subtraction, and a constant. It follows that the relation is primitive recursive. It follows immediately that the relation > is primitive recursive since its characteristic function is χ > (n, m) = 1 + χ (n, m), which is a primitive recursive function since it is obtained by substitution using the primitive recursive functions χ (n, m), cut-off subtraction and a constant. Let χ (n, m) = 1 if n m, χ (n, m) = 0 if n < m. Then we can write: χ (n, m) = sg(n + m) + χ eq (n, m) Therefore the characteristic function of the relation is a primitive recursive function, since it is obtained by substitution using the primitive recursive functions sg, cut-off subtraction, add, and χ eq. It follows that the relation is primitive recursive. It follows immediately that the relation < is primitive recursive since its characteristic function is equal to 1 + χ (n, m), which is a primitive recursive function since it is obtained by substitution using the primitive recursive functions χ (n, m), cut-off subtraction and a constant. 17. The function C(n, r) = n C r (the binomial coefficient) if r n, C(n, r) = 0 if r > n, is primitive recursive. Proof: We have: nc r = n!/(r!(n - r)!) = fac(n)/(fac(r)fac(n - r)) = quot(fac(n), fac(r)fac(n + r)) Note that for r > n this reduces to quot(fac(n), fac(r)) = 0. Therefore we can write C(n, r) as a single equation as: C(n, r) = quot(fac(n), fac(r)fac(n + r)) It follows that C(n, r) is a primitive recursive function since it is obtained by substitution using the primitive recursive functions quot, fac, mult, cut-off subtraction and projection functions (projection functions not shown explicitly). 18. The function div(n, y) = 1 if y n, div(n, y) = 0 otherwise, is a primitive recursive function. Proof: If y n we have rem(n, y) = 0. Therefore we can write: div(n, y) = 1 + sg(rem(n, y)) It follows that div(n, y) is primitive recursive, since it is obtained by substitution using the primitive recursive functions sg, rem, cut-off subtraction and a constant. 19. Suppose that the function f: N² N is primitive recursive and that a is a fixed natural number. Then each of the following functions of two variables, g 1, g 2, g 3, is primitive recursive: g 1 (n 1, n 2 ) = f(a, n 1, n 2 ) g 2 (n 1, n 2 ) = f(n 1, a, n 2 ) g 3 (n 1, n 2 ) = f(n 1, n 2, a) Proof: In the case of g 1 we can write: g 1 (n 1, n 2 ) = f(c a ²(n 1, n 2 ), U 1 ²(n 1, n 2 ), U 2 ²(n 1, n 2 )) Thus, g 1 is primitive recursive because it is defined by substitution using the primitive recursive functions f, C a ², U 1 ², and U 2 ². Analogous arguments show that g 2 and g 3 are also primitive recursive.

11 20. If the function f: N² N is primitive recursive, then so is the function g: N N given by g(n) = f(n, n). Proof: We can write: g(n) = f(u 1 ¹(n), U 1 ¹(n)) Thus, g(n) can be defined by substitution in terms of the primitive recusrive functions f and U 1 ¹, so it must itself be a primitive recursive function. 21. Proof: The function g satisfies the equations: g(n 1, n 2,..., n k, 0) = 0 g(n 1, n 2,..., n k, z + 1) = g(n 1, n 2,..., n k, z) + f(n 1, n 2,..., n k, z + 1) This shows that g is defined by primitive recursion from the primitive recursive functions add and f, using constants. It follows that g is itself primitive recursive. 22. The set Pr of prime numbers is primitive recursive. Proof: A natural number n > 0 is prime if and only if it has exactly two divisors, itself and 1. Thus n > 0 is prime if and only if [y=1, n] div(n, y) = 2. The characteristic function of the set Pr of prime numbers can thus be written as: χ Pr (n) = χ eq ( [y=1, n] div(n, y), 2) Now let g: N² N be the function defined by: g(n, z) = 0 if z = 0 g(n, z) = [y=1, z] div(n, y) otherwise Since div is a primitive recursive function, and g is a bounded summation function of the form appearing in 21 above, it follows that g is a primitive recursive function. Also, for n > 0, we have: g(n, n) = [y=1, n] div(n, y) Now let h: N N be the function given by h(n) = g(n, n), which is primitive recursive by item 20 above. We observe that for all natural numbers n (i.e., n = 0 and n > 0), we have: χ Pr (n) = χ eq (h(n), 2) Thus χ Pr is obtained by substitution from the primitive recursive functions χ eq and h using constants, and so is a primitive recursive function. It follows that the set of prime numbers Pr is a primitive recursive set. 23. The set E of even numbers is primitive recursive. Proof: Let h(n) χ E (n) (the characteristic function of E). Then we have: h(0) = 1 h(n + 1) = g(n, h(n)) = 1 + h(n)

12 Thus χ E is obtained by substitution and primitive recursion using cut-off subtraction (a primitive recursive function), constants and projection functions (not shown explicitly), and is therefore primitive recursive. It follows that the set E of even numbers is primitive recursive. 24. Proof: Let R 3 be the two-place relation defined by: R 3 (n, m) not R 1 (n, m) and not R 2 (n, m) Then it follows from the 'definition by cases' theorem for relations that f is primitive recursive provided that R 1, R 2 and R 3 are primitive recursive relations which are mutually exclusive and exhaustive. From the definition of R 3, at least one of the relations holds for all pairs (n, m), so the three relations are exhaustive. To show that the three relations are mutually exclusive, note first that R 3 is mutually exclusive with both R 1 and R 2 by construction. Next, we note that if max(n, 2m) 21, then either n 21 3n + 4m 63, or 2m 21 m 11 3n + 4m 44. Thus R 1 and R 2 are mutually exclusive. Therefore the three relations are mutually exclusive and exhaustive. Next we show that each of the relations is primitive recursive. The characteristic function of R 1 is: χ R1 (n, m) = χ (21, max(n, 2m)) Thus, χ R1 is obtained by substitution from the primitive recursive functions χ, max and mult using constants. It follows that χ R1 is a primitive recursive function, and hence R 1 is a primitive recursive relation. The characteristic function of R 2 is: χ R2 (n, m) = χ eq (3n + 4m, 43) Thus χ R2 is obtained by substitution from the primitive recursive functions χ eq, add and mult using constants. It follows that χ R2 is a primitive recursive function, and hence that R 2 is a primitive recursive relation. Finally, the characteristic function of R 3 is: χ R3 (n, m) = 1 + sg(χ R1 (n, m) + χ R2 (n, m)) Thus χ R3 is obtained by substitution from the primitive recursive functions χ R1, χ R2, sg, add and cut-off subtraction using constants. It follows that χ R3 is a primitive recursive function, and hence that R 3 is a primitive recursive relation. Thus f is a primitive recursive function. 25. Let the function g: N² N be defined by g(n, z) = μy z ( n - y² = 0), and let the function h: N N be defined by h(n) = g(n, n). Then h is primitive recursive. Proof: By item 20 above, h is primitive recursive if g is. By the key theorem on bounded minimization on a function, g is primitive recursive if the function f: N² N given by f(n, y) = n - y² is primitive recursive. But f is clearly primitive recursive since it is obtained by substitution from the primitive recursive functions adf and mult. Hence g, and therefore h, are primitive recursive.

13 26. The function p which enumerates the prime numbers is primitive recursive. Proof: The function p is defined so that p(0) = 1, p(1) = 2 (the first prime number), and for n > 1, p(n) is the nth prime number. Having a specified value for p(0), the natural way to define p by recursion is to put p(n + 1) = the smallest natural number y such that y is prime and p(n) < y. We can write this mathematically as: p(n + 1) = μy (χ Pr (y) = 1 and p(n) < y) Now define a two-place relation S as follows: S(m, y) χ Pr (y) = 1 We have: χ S (m, y) = χ eq (χ Pr (y), 1) Therefore χ S is primitive recursive since it is obtained by substitution from the primitive recursive functions χ eq and χ Pr using constants. Thus the relation S is primitive recursive. Also the relation < is primitive recursive. Now define the relation R by: R(m, y) S(m, y) and m < y χ Pr (y) = 1 and m < y We have: χ R (m, y) = χ S (m, y) χ > (y, m) Therefore χ R is primitive recursive since it is obtained by substitution from the primitive recursive functions mult, χ S and χ >. It follows from the key theorem on bounded minimization on a relation that the function g: N² N given by g(m, z) = μy z (χ Pr (y) = 1 and m < y) is primitive recursive. We observe that g(p(n), z) = p(n + 1) provided that p(n + 1) z. We need to find a suitable bound z. We can use a standard result from number theory that the nth prime number p(n) satisfies the inequality p(n) exp(2, exp(2, n-1)) where exp refers to exponentiation. Therefore let h:n N be defined by h(n) = exp(2, exp(2, n)). Then h is a primitive recursive function since it is obtained by substitution from the primitive recursive function exp using constants. We also have p(n + 1) h(n). It follows that p(n + 1) = g(p(n), h(n)) where g and h are primitive recursive. Thus we can define p by primitive recursion as p(0) = 1 p(n + 1) = g(p(n), h(n)) using the primitive recursive functions g, h and the constant 1. It follows that p is a primitive recursive function. 27. Let f:n N be the function defined by f(n) = μy (n < 3ʸ). Then f is primitive recursive. Proof: We use the fact that n < 3ʸ for all n when y = n. Therefore we must have f(n) n. Define the function g: N² N by g(n, z) = μy z (n < 3ʸ). Then we have f(n) = g(n, n), i.e., f equals g when the bound z in g is set equal to n. Therefore f(n) is primitive recursive if g(n, n) is primitive recursive by item 20 above. By the key theorem on bounded minimization on a relation, g(n, n) is primitive recursive if the relation R(n, y) n < 3ʸ is primitive recursive. But the characteristic function of R can be written as χ R (n, y) = χ > (exp(3, y), n) which is defined by substitution using the primitive recursive functions χ >, exp and constants. Thus R is a primitive recursive relation, so g and therefore f are primitive recursive functions. 28. Let f:n N be the function defined by f(0) = 1 f(n + 1) = μy (f(n) < y and y is not divisible by 4) Then f is primitive recursive.

14 Proof: We can use the approach of showing that f is defined by primitive recursion from known primitive recursive functions. In doing this it is the definition of f(n + 1) which requires attention, since f(0) is simply defined using a constant. Let R(m, y) be the relation given by: R(m, y) m < y and y is not divisible by 4 The characteristic function of R is χ R (m, y) = χ > (y, m) (1 + χ eq (div(y, 4), 0)) Thus χ R is obtained by substitution from the primitive recursive functions χ >, χ eq, mult, div, and cut-off subtraction using constants. It follows that χ R is a primitive recursive function, and hence that R is a primitive recursive relation. Hence, by the key theorem on bounded minimization on a relation, the function g: N² N given by g(m, z) = μy z R(m, y) is primitive recursive. We have f(n + 1) = μy R(f(n), y) and so to show that f(n + 1) is given by a primitive recursive function we require a bound z on y. To find a suitable bound, note that the smallest possible value for f(n + 1) is f(n) + 1. However, if f(n) + 1 is divisible by 4, then the next smallest is f(n) + 2. Thus, the biggest gap that can occur between f(n) and f(n + 1) is 2. Therefore a suitable bound z on y is f(n) + 2. So letting h: N N be defined by h(m) = m + 2 (so that h is primitive recursive), we have f(n + 1) = μy h(f(n)) R(f(n), y) = g(f(n), h(f(n))) Therefore we have f(0) = 1 f(n + 1) = g(f(n), h(f(n))) so that f is obtained by primitive recursion from the constant 1 and the primitive recursive functions g and h. Thus f is a primitive recursive function.