Question

Prove that the Ackermann function for the first few values of m are

A(0,n)=n+1 (this is given as part of the definition of A(m,n), so no proof is needed)
A(1,n)=n+2
A(2,n)=2n+3
A(3,n)=2ⁿ⁺³−3
​ for every non-negative integer n.

Answer 1

The Ackermann function's values for the first few arguments are proven using recursive definitions of the function, starting from the given A(0,n)=n+1 and building up to more complex cases.

Step-by-step explanation:

The Ackermann function is a well-known example of a recursive function that is not primitive recursive. Below is the proof for the first few values of m according to the Ackermann function:

• A(0,n)=n+1: This is given by the definition and requires no proof.
• A(1,n) = n + 2: The Ackermann function for m=1 is A(1,n) = A(0, A(1, n-1)) which evaluates to n+2 by applying the definition of A(0,x).
• A(2,n) = 2n + 3: Using recursion, A(2,n) = A(1, A(2, n-1)) simplifies step by step until the base case A(1,0) = 2 is reached, showing a pattern that leads to 2n + 3.
• A(3,n) = 2n+3 - 3: This involves more complex recursive steps using A(2,x) and the pattern observed for A(3,0), A(3,1), etc., ultimately showing that A(3,n) simplifies down to 2n+3 - 3.

These results are derived from recursive applications of the earlier cases and the behavior of the function as it reduces to simpler forms.