Given the functions:
Predecessor function: $$(\operatorname{Pred}(x)=x-1)$$​​
and $$(T(x,y+1)=\operatorname{pred}(T(x,y)),\ T(x,0)=x.)$$​​
Which of the following operations is a candidate for T?

Correct option is A

Given:
​$$\operatorname{Pred}(x)=x-1$$​​
and the recursive definition:
​$$T(x,y+1)=\operatorname{Pred}(T(x,y))$$​​
with the base condition:
T(x, 0) = x
We need to identify the operation represented by T(x, y).
Step-by-Step Solution
Starting with the base case:
T(x, 0) = x
Now put y = 0 in the recursive equation:
​$$T(x,1)=\operatorname{Pred}(T(x,0))$$​​
Since T(x, 0) = x:
​$$T(x,1)=\operatorname{Pred}(x)=x-1$$​​
Again:
​$$T(x,2)=\operatorname{Pred}(T(x,1))$$​​
= x − 2
Similarly:
T(x, 3) = x - 3
Therefore, in general:
​$$\boxed{T(x,y)=x-y}$$​​
So, T represents the subtraction operation.
Verification
For example, let x = 7 and y = 3:
T(7, 0) = 7
T(7, 1) = 6
T(7, 2) = 5
T(7, 3) = 4
Thus:
T(7, 3) = 7 – 3 = 4
Hence:
​$$\boxed{\text{(a) Subtraction Operation}}$$​​
Information Booster
1. The predecessor function decreases its argument by one:
​$$\operatorname{Pred}(x)=x-1$$​​
2. The recurrence applies the predecessor function once for every increment in y.
3. Therefore:
T(x, y) = x - y
4. The base case T(x, 0) = x confirms that no subtraction occurs when y = 0.
5. This is an example of defining an arithmetic operation recursively using a predecessor function.