Prove that the outer loop for this algorithm is correct, using the following steps. You may assume for simplicity that the inner while loop is correct (i.e., it inserts the element A[j] in its proper position of the sub-array A[1..j]). Given the following pre-condition, loop invariant, and post-condition: • Pre-condition: A[1..n] is an array of integers. • Loop invariant: at the start of each iteration, the sub-array A[1..j – 1] consists of the elements originally in A[1..j – 1], but in sorted order. • Post-condition: the array A[1..n] consists of the elements originally in A[1..n], but in sorted order. Prove the following: • Initialization: If the pre-condition is true, then the loop invariant is initially true (prior to the first iteration of the loop). • Maintenance: If the loop invariant is true before an iteration, it remain true before the next iteration. • Termination: The loop will eventually terminate. And when it does, the loop invariant implies the post-condition. 1: for j=2: A.length do key = A[j] i = j – 1 while i > 0 and A[i] > key do A[i + 1] = A[i] i = i – 1 A[i + 1] = key 2: 3: 4: 5: 6: 7: