Answer:
Given Loop Variant P = a[0], a[1] ... a[i]
It is product of n terms in array
Explanation:
The Basic Step: i = 0, loop invariant p=a[0], it is true because of 'p' initialized as a[0].
Induction Step: Assume that for i = n - 3, loop invariant p is product of a[0], a[1], a[2] .... a[n - 3].
So, after that multiply a[n - 2] with p, i.e P = a[0], a[1], a[2] .... a[n - 3].a[n - 2].
After execution of while loop, loop variant p becomes: P = a[0], a[1], a[2] .... a[n -3].a[n -2].
for the case i = n-2, invariant p is product of a[0], a[1], a[2] .... a[n-3].a[n-2]. It is the product of (n-1) terms. In while loop, incrementing the value of i, so i=n-1
And multiply a[n-1] with p, i.e P = a[0].a[1].a[2].... a[n-2].
a[n-1]. i.e. P=P.a[n-1]
By the assumption for i=n-3 loop invariant is true, therefore for i=n-2 also it is true.
By induction method proved that for all n > = 1 Code will return product of n array elements.
While loop check the condition i < n - 1. therefore the conditional statement is n - i > 1
If i = n , n - i = 0 , it will violate condition of while loop, so, the while loop will terminate at i = n at this time loop invariant P = a[0].a[1].a[2]....a[n-2].a[n-1]
