1. Primitive Recursive Predicates
Theorem 5.4: Let C be a PRC class.
Let the functions g, h and the predicate P belong to C .
Let
f(x1, …, xn) = g(x1, …, xn) if P(x1, …, xn)
= h(x1, …, xn) otherwise.
Then f belongs to C .
Proof: f obviously belongs to C , because it can be written as:
f(x1, …, xn) = g(x1, …, xn)P(x1, …, xn) h(x1, …, xn)(P(x1, …, xn))
October 3, 2017
Theory of Computation Lecture 9: Primitive Recursive Functions IV

2. Primitive Recursive Predicates
Corollary 5.5: Let C be a PRC class.
Let n-ary functions g1, …, gm, h and predicates P1, …, Pm belong to C , and let
Pi(x1, …, xn) & Pj(x1, …, xn) = 0 for all 1  i < j  m and all x1, …, xn. If
f(x1, …, xn) = g1(x1, …, xn) if P1(x1, …, xn)
= g2(x1, …, xn) if P2(x1, …, xn)
: :
= gm(x1, …, xn) if Pm(x1, …, xn)
= h(x1, …, xn) otherwise.
Then f belongs to C .
October 3, 2017
Theory of Computation Lecture 9: Primitive Recursive Functions IV

3. Primitive Recursive Predicates
Proof: The idea is to use induction on the variable m.
Theorem 5.4 provides the case for m = 1, so we just have to do the inductive step. Let
f(x1, …, xn) = g1(x1, …, xn) if P1(x1, …, xn)
: :
= gm+1(x1, …, xn) if Pm+1(x1, …, xn)
= h(x1, …, xn) otherwise, and let
h’(x1, …, xn) = gm+1(x1, …, xn) if Pm+1(x1, …, xn)
= h(x1, …, xn) otherwise. Then
= gm(x1, …, xn) if Pm(x1, …, xn)
= h’(x1, …, xn) otherwise.
October 3, 2017
Theory of Computation Lecture 9: Primitive Recursive Functions IV

4. Primitive Recursive Predicates
The previous steps showed that
f belongs to C for m = 1
whenever f belongs to C for a particular m, then f also belongs to C for m + 1.
Conclusion: f belongs to C for any natural number m.
October 3, 2017
Theory of Computation Lecture 9: Primitive Recursive Functions IV

5. Primitive Recursive Predicates
Theorem 6.1: Let C be a PRC class.
If f(t, x1, …, xn) 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.
October 3, 2017
Theory of Computation Lecture 9: Primitive Recursive Functions IV

6. Primitive Recursive Predicates
Proof I:
can be obtained by recursion in the following way:
g(0, x1, …, xn) = f(0, x1, …, xn)
g(t + 1, x1, …, xn) = g(t, x1, …, xn) + f(t + 1, x1, …, xn)
Since + is primitive recursive, g belongs to C .
October 3, 2017
Theory of Computation Lecture 9: Primitive Recursive Functions IV

7. Primitive Recursive Predicates
Proof II:
can be obtained by recursion in the following way:
h(0, x1, …, xn) = f(0, x1, …, xn)
h(t + 1, x1, …, xn) = h(t, x1, …, xn)  f(t + 1, x1, …, xn)
Since  is primitive recursive, h belongs to C .
October 3, 2017
Theory of Computation Lecture 9: Primitive Recursive Functions IV

8. 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, x1, …, xn) = 0
h(0, x1, …, xn) = 1
October 3, 2017
Theory of Computation Lecture 9: Primitive Recursive Functions IV

9. Primitive Recursive Predicates
As a “by-product” of the preceding idea we obtained the following corollary:
Corollary 6.2: If f(t, x1, …, xn) belongs to the PRC class C , then so do the following two functions:
October 3, 2017
Theory of Computation Lecture 9: Primitive Recursive Functions IV

10. Primitive Recursive Predicates
Theorem 6.3: If the predicate P(t, x1, …, xn) belongs to some PRC class C, then so do the following two predicates:
(t)y P(t, x1, …, xn) and (t)y P(t, x1, …, xn)
Proof:
October 3, 2017
Theory of Computation Lecture 9: Primitive Recursive Functions IV

11. Primitive Recursive Predicates
Sometimes we may want to use the quantifiers
(t)<y P(t, x1, …, xn) and (t)<y P(t, x1, …, xn).
Then Theorem 6.3 is still valid, which is obvious from the following two relations:
(t)<y P(t, x1, …, xn)  (t)y [t = y  P(t, x1, …, xn)]
(t)<y P(t, x1, …, xn)  (t)y [t  y & P(t, x1, …, xn)]
October 3, 2017
Theory of Computation Lecture 9: Primitive Recursive Functions IV

12. Primitive Recursive Predicates
Example 12: y | x
y | x means “y is a divisor of x.”
For example,
4 | 15 is
false.
3 | 9 is
true.
This predicate is primitive recursive because
y | x  (t)x (y  t = x).
October 3, 2017
Theory of Computation Lecture 9: Primitive Recursive Functions IV

13. 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)].
October 3, 2017
Theory of Computation Lecture 9: Primitive Recursive Functions IV

14. Theory of Computation Lecture 9: Primitive Recursive Functions IV
Minimalization
Let the predicate P(t, x1, …, xn) 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.)
October 3, 2017
Theory of Computation Lecture 9: Primitive Recursive Functions IV

15. Theory of Computation Lecture 9: Primitive Recursive Functions IV
Minimalization
Let us take a closer look at the function
Let us assume that there is a value t0 that is the smallest value of t  y for which P(t, x1, …, xn) is true:
P(t, x1, …, xn) = 0 for t < t0
P(t0, x1, …, xn) = 1. Then
Hence,
so that g(y, x1, …, xn) is the least value of t for which P(t, x1, …, xn) is true.
October 3, 2017
Theory of Computation Lecture 9: Primitive Recursive Functions IV

16. Theory of Computation Lecture 9: Primitive Recursive Functions IV
Minimalization
Then we define:
Thus, minty P(t, x1, …, xn) is the smallest value of ty for which P(t, x1, …, xn) 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, x1, …, xn) belongs to some PRC Class C and f(y, x1, …, xn) = minty P(t, x1, …, xn), then f also belongs to C .
We will call the operation “minty” bounded minimalization.
October 3, 2017
Theory of Computation Lecture 9: Primitive Recursive Functions IV

17. Theory of Computation Lecture 9: Primitive Recursive Functions IV
Minimalization
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 = mintx [(t + 1)  y > x]
Note that this equation gives us x/0 = 0.
October 3, 2017
Theory of Computation Lecture 9: Primitive Recursive Functions IV

18. Theory of Computation Lecture 9: Primitive Recursive Functions IV
Minimalization
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.
October 3, 2017
Theory of Computation Lecture 9: Primitive Recursive Functions IV

19. Theory of Computation Lecture 9: Primitive Recursive Functions IV
Minimalization
Example 16: pn
Here, for n > 0, pn is the n-th prime number (in order of size).
In order to make pn a total function, we set p0 = 0.
Then we have:
p0 =
p1 = 2
p2 = 3
p3 = 5
p4 = 7
October 3, 2017
Theory of Computation Lecture 9: Primitive Recursive Functions IV

20. Theory of Computation Lecture 9: Primitive Recursive Functions IV
Minimalization
Now consider the following recursion equations:
p0 = 0
pn+1 = mintpn!+1[Prime(t) & t > pn].
To see that these equations are correct, we must verify the following inequality:
pn+1  pn! + 1.
October 3, 2017
Theory of Computation Lecture 9: Primitive Recursive Functions IV

21. Theory of Computation Lecture 9: Primitive Recursive Functions IV
Minimalization
pn+1  pn! + 1.
Note that for 0 < i  n we have:
(pn! + 1)/pi = pn!/pi + 1/pi = K + 1/pi , where K is an integer.
Why is pn! always divisible by pi ?
Example: n = 4, i = 2
Then pn! = 1234567, and pi = 3, so pn!/pi = 124567 = K.
In other words, pi is always one of the factors in pn!.
According to the equation (pn! + 1)/pi = K + 1/pi , pn! + 1 is not divisible by any of the primes p1, …, pn.
October 3, 2017
Theory of Computation Lecture 9: Primitive Recursive Functions IV

22. Theory of Computation Lecture 9: Primitive Recursive Functions IV
Minimalization
So either pn! + 1 is itself a prime or it is divisible by a prime > pn.
In either case there is a prime q such that pn < q  pn! + 1, which gives us the inequality that we wanted to verify:
pn+1  pn! + 1.
October 3, 2017
Theory of Computation Lecture 9: Primitive Recursive Functions IV