**Title: Proof by Induction for Elements in an SList of Depth p****Objective:** Use mathematical induction to demonstrate that the number of elements in any SList (Skip List) of depth \( p \) is \( 2^p \).**Introduction:**In computer science, a Skip List is a data structure that allows for efficient search, insertion, and deletion operations. The Skip List can be seen as a layered, linked list structure with multiple levels. In this exercise, we aim to prove that the number of elements in a Skip List of depth \( p \) is \( 2^p \).**Proof by Induction:**1. **Base Case:** - Consider the simplest Skip List of depth \( p = 0 \). - At this level, there is just one element. - Therefore, the number of elements \( = 2^0 = 1 \). - Thus, the base case holds.2. **Inductive Step:** - Assume that for a Skip List of depth \( k \), the number of elements is \( 2^k \). - Now, consider a Skip List of depth \( k+1 \). - At this new level, each element on level \( k \) is linked to two additional elements in level \( k+1 \). - Thus, the number of elements at level \( k+1 \) is twice the number at level \( k \), hence \( 2 \times 2^k = 2^{k+1} \). - Thus, by the principle of mathematical induction, the number of elements in any Skip List of depth \( p \) is \( 2^p \).**Conclusion:**Through our inductive proof, we conclusively show that a Skip List of depth \( p \) contains \( 2^p \) elements, corroborating the efficiency and scalability of Skip Lists in data organization.

We have to prove that the number of elements in any SList of depth p is . Notice that the definition of SList states that, B. x where , the real numbers. R. where X and Y are SLists having the same number of elements, and the last number in X is less than the first number in Y. The number p counts the depth of parentheses of the SList (the depth of the SList) and every number in the list will be inside p pairs of parentheses. We will use the induction on the depth of parentheses, p to prove the result. Base Case: If, the list L contains only a single number and it can represented as. Inductive Hypothesis: Suppose an inductive hypothesis that any SList of depth, has elements for some . Inductive Step: Given an SList L of depth k, we know that the list can be represented as for some X and Y of depth. By induction hypothesis both X and Y has elements. The number of elements in L is the sum of the elements in X and Y. So, we have: Therefore, the number of elements in any SList of depth p is. ..