1. October 1, 2009Theory of Computation Lecture 8: Primitive Recursive Functions IV 1 Primitive Recursive Predicates Theorem 6.1: Let C be a PRC class. If f(t, x 1, …, x n ) belongs to C, then so do the functions To prove this theorem, we will show that g and h can be obtained from f by primitive recursion.

2. October 1, 2009Theory of Computation Lecture 8: Primitive Recursive Functions IV 2 Primitive Recursive Predicates Proof I: can be obtained by recursion in the following way: g(0, x 1, …, x n ) = f(0, x 1, …, x n ) g(t + 1, x 1, …, x n ) = g(t, x 1, …, x n ) + f(t + 1, x 1, …, x n ) Since + is primitive recursive, g belongs to C.

3. October 1, 2009Theory of Computation Lecture 8: Primitive Recursive Functions IV 3 Primitive Recursive Predicates Proof II: can be obtained by recursion in the following way: h(0, x 1, …, x n ) = f(0, x 1, …, x n ) h(t + 1, x 1, …, x n ) = h(t, x 1, …, x n )  f(t + 1, x 1, …, x n ) Since  is primitive recursive, h belongs to C.

4. October 1, 2009Theory of Computation Lecture 8: Primitive Recursive Functions IV 4 Primitive Recursive Predicates In some cases we might want to begin the iteration at 1 instead of 0: Then we just need to replace the initial recursion equations: g(0, x 1, …, x n ) = 0 h(0, x 1, …, x n ) = 1

5. October 1, 2009Theory of Computation Lecture 8: Primitive Recursive Functions IV 5 Primitive Recursive Predicates As a “by-product” of the preceding idea we obtained the following corollary: Corollary 6.2: If f(t, x 1, …, x n ) belongs to the PRC class C, then so do the following two functions:

6. October 1, 2009Theory of Computation Lecture 8: Primitive Recursive Functions IV 6 Primitive Recursive Predicates Theorem 6.3: If the predicate P(t, x 1, …, x n ) belongs to some PRC class C, then so do the following two predicates: (  t)  y P(t, x 1, …, x n ) and (  t)  y P(t, x 1, …, x n ) Proof:

7. October 1, 2009Theory of Computation Lecture 8: Primitive Recursive Functions IV 7 Primitive Recursive Predicates Sometimes we may want to use the quantifiers (  t) <y P(t, x 1, …, x n ) and (  t) <y P(t, x 1, …, x n ). Then Theorem 6.3 is still valid, which is obvious from the following two relations: (  t) <y P(t, x 1, …, x n )  (  t)  y [t = y  P(t, x 1, …, x n )] (  t) <y P(t, x 1, …, x n )  (  t)  y [t  y & P(t, x 1, …, x n )]

8. October 1, 2009Theory of Computation Lecture 8: Primitive Recursive Functions IV 8 Primitive Recursive Predicates Example 12: y | x y | x means “y is a divisor of x.” For example, 4 | 15 is This predicate is primitive recursive because y | x  (  t)  x (y  t = x). false. 3 | 9 is true.

9. October 1, 2009Theory of Computation Lecture 8: Primitive Recursive Functions IV 9 Primitive Recursive Predicates Example 13: Prime(x) Prime(x) is the predicate “x is a prime.” It is primitive recursive since Prime(x)  x > 1 & (  t)  x [t = 1  t = x   (t | x)].

10. October 1, 2009Theory of Computation Lecture 8: Primitive Recursive Functions IV 10Minimalization Let the predicate P(t, x 1, …, x n ) belong to some PRC class C. Then by Theorem 6.1 the function also belongs to C. (Remember the primitive recursive “negation” function  we defined earlier.)

11. October 1, 2009Theory of Computation Lecture 8: Primitive Recursive Functions IV 11Minimalization Let us take a closer look at the function Let us assume that there is a value t 0 that is the smallest value of t  y for which P(t, x 1, …, x n ) is true: P(t, x 1, …, x n ) = 0 for t < t 0 P(t 0, x 1, …, x n ) = 1. Then Hence, so that g(y, x 1, …, x n ) is the least value of t for which P(t, x 1, …, x n ) is true.

12. October 1, 2009Theory of Computation Lecture 8: Primitive Recursive Functions IV 12Minimalization Then we define: Thus, min t  y P(t, x 1, …, x n ) is the smallest value of t  y for which P(t, x 1, …, x n ) is true, if such t  y exists, otherwise it is 0. Using Theorems 5.4 and 6.3, we have: Theorem 7.1: If P(t, x 1, …, x n ) belongs to some PRC Class C and f(y, x 1, …, x n ) = min t  y P(t, x 1, …, x n ), then f also belongs to C. We will call the operation “min t  y ” bounded minimalization.

13. October 1, 2009Theory of Computation Lecture 8: Primitive Recursive Functions IV 13Minimalization Example 14:  x/y    is the floor function. For example,  8/3  = 2. So  x/y  is the integer part of the quotient x/y.  x/y  is primitive recursive as shown by the equation  x/y  = min t  y [(t + 1)  y > x] Note that this equation gives us  x/0  = 0.

14. October 1, 2009Theory of Computation Lecture 8: Primitive Recursive Functions IV 14Minimalization Example 15: R(x, y) R(x, y) is the remainder when x is divided by y. We can also write R(x, y) = x mod y (“modulo”). Obviously, it is true that x/y =  x/y  + R(x, y)/y Therefore, we can write: R(x, y) = x -  (y   x/y  ) This shows that R(x, y) is primitive recursive. Note that R(x, 0) = x.

15. October 1, 2009Theory of Computation Lecture 8: Primitive Recursive Functions IV 15Minimalization Example 16: p n Here, for n > 0, p n is the n-th prime number (in order of size). In order to make p n a total function, we set p 0 = 0. Then we have: p0 =p0 =p0 =p0 = 0 p1 =p1 =p1 =p1 =2 p2 =p2 =p2 =p2 =3 p3 =p3 =p3 =p3 =5 p4 =p4 =p4 =p4 =7

16. October 1, 2009Theory of Computation Lecture 8: Primitive Recursive Functions IV 16Minimalization Now consider the following recursion equations: p 0 = 0 p n+1 = min t  p n !+1 [Prime(t) & t > p n ]. To see that these equations are correct, we must verify the following inequality: p n+1  p n ! + 1.

17. October 1, 2009Theory of Computation Lecture 8: Primitive Recursive Functions IV 17Minimalization p n+1  p n ! + 1. Note that for 0 < i  n we have: (p n ! + 1)/p i = p n !/p i + 1/p i = K + 1/p i, where K is an integer. Why is p n ! always divisible by p i ? Example: n = 4, i = 2 Then p n ! = 1  2  3  4  5  6  7, and p i = 3, so p n !/p i = 1  2  4  5  6  7 = K. In other words, p i is always one of the factors in p n !. According to the equation (p n ! + 1)/p i = K + 1/p i, p n ! + 1 is not divisible by any of the primes p 1, …, p n.

18. October 1, 2009Theory of Computation Lecture 8: Primitive Recursive Functions IV 18Minimalization So either p n ! + 1 is itself a prime or it is divisible by a prime > p n. In either case there is a prime q such that p n < q  p n ! + 1, which gives us the inequality that we wanted to verify: p n+1  p n ! + 1.