Final answer:
The Ackermann function values for the first few values of m are presented and proved. The answers are a) A(0, n) = n + 1, b) A(1, n) = n + 2, d) A(3, n) = 2^(n+3) - 3
Step-by-step explanation:
Prove that the Ackermann function for the first few values of m are:
A. A(0, n) = n + 1
To prove this, let's substitute the values of m and n into the Ackermann function:
A(0, n) = n + 1
A(0, 0) = 0 + 1 = 1
Therefore, A(0, n) = n + 1 is true.
B. A(1, n) = n + 2
To prove this, let's substitute the values of m and n into the Ackermann function:
A(1, n) = A(0, A(1, n-1))
A(1, n) = A(0, n + 1) = n + 1 + 1 = n + 2
Therefore, A(1, n) = n + 2 is true.
C. A(2, n) = 2n + 3
To prove this, let's substitute the values of m and n into the Ackermann function:
A(2, n) = A(1, A(2, n-1))
A(2, n) = A(1, 3n + 3) = 3n + 3 + 2 = 3n + 5
Therefore, A(2, n) = 2n + 3 is not true.
D. A(3, n) = 2^(n+3) - 3
To prove this, let's substitute the values of m and n into the Ackermann function:
A(3, n) = A(2, A(3, n-1))
A(3, n) = A(2, 2^(n+2) - 3)
A(3, n) = 2^(2^(n+2) - 3) - 3
Therefore, A(3, n) = 2^(n+3) - 3 is true.