Question

Доказать что f(x)=2 приметивно-рекурсивная функция

Answer 1

f(x,y)=2^(x^2+y)+y^x!

f(x,y)=2^(x^2+y

f(x,0)=2^(x^2)

f(x,y+1)=2*2^(x^2+y)=h(y(f(y))

f(x,y)=y^x!

u(y,x)=y^x - примитинво рекурсивна c(x)=x! -примитивно рекрсивна, значчит u(y,c(x)) перимитивно рекурсивна а значит вся функция прмиимтивно рекурсивна

Answer 2

To prove that a function f(x) is primitive-recursive, we need to show that it can be constructed using the primitive-recursive operations: zero function, successor function, projection, and composition, through a finite number of steps. In this case, we want to show that the constant function f(x) = 2 is primitive-recursive.

1. Zero function: The zero function z(x) is primitive-recursive, defined as z(x) = 0 for all x.

2. Successor function: The successor function s(x) is primitive-recursive, defined as s(x) = x + 1.

3. Constant function: The constant function with a value k is primitive-recursive. In this case, f(x) = 2 is a constant function, so it's primitive-recursive.

Now, let's go through the steps of constructing the constant function f(x) = 2 using the primitive-recursive operations:

Step 1: Use the zero function to get the constant 0: g(x) = z(x) = 0.

Step 2: Use the successor function to get the constant 1: h(x) = s(g(x)) = s(0) = 1.

Step 3: Use the successor function again to get the constant 2: f(x) = s(h(x)) = s(1) = 2.

Since we can construct the constant function f(x) = 2 using a finite number of steps with the primitive-recursive operations, we have proven that f(x) = 2 is primitive-recursive.