First, if the students have any questions answer them.

Go over exercise 4 of chapter 2 of the textbook (page 81) slowly and
make sure the students understand every step. 
Here is the solution:

Termination: Let E = l + 1 - i. E is definitely an integer because l and i
  are both integers. Sine f <= l by the precondition: E_0 = l + 1 - i_0 =
  l + 1 - f >= 1, and since i_k <= l at the begining of each iteration of the
  loop, i_{k+1} = i_k + 1, i_{k+1} <= l + 1, so E_{k+1} = l + 1 - i_{k+1} >=0
  So E_k is always a natural number. Furthermore, if there is an iteration 
  number k + 1 of the loop, then
            E_{k+1} = l + 1 - i_{k+1} = l + 1 - (i_k + 1) 
                    = l + 1 - i_k - 1 = E_k - 1 
                    < E_k
  Therefore, E_0 > E_1 > ...  is a decreasing sequence of natural numbers and 
  is finite, which implies that the loop terminates.

Partial correctness: Under the assumption that the precondition holds, we 
  prove the following loop invariant (LI) by induction on the number of 
  iterations performed:
     "i <= l + 1 and t is the number of ccurences of x in A[f...(i - 1)]".
  Base: t_0 = t = 0 and A[f...(i_0 - 1)] = A[f...(f - 1)] which is the
        empty array by convention. Also, i_0 = f <= l < l + 1 by the 
        precondition. 
  Ind. Step: Assuming that LI_k is true, we want to prove that LI_{k+1} 
        holds too.
        Case 1: If there is no iteration number k + 1, then LI_{k+1} is
           equivalent to LI_k so it is true by the Ind. Hyp.
        Case 2: If there is an iteration number k + 1, then that means
           i_k <= l (by the loop test), so i_{k+1} = i_k + 1 <= l + 1.
           Now we consider two subcases:
           Subcase A: If A[i_k] =/= x, then the test "A[i] = x" fails and
              t_{k+1} = t_k. Since t_k is the number of occurences of x in
              A[f...(i_k - 1)] by the Ind. Hyp. and A[i_k] =/= x, 
              t_{k+1} = t_k is the number of occurences of x in A[f...i_k]
              which is the same as A[f...(i_{k+1} - 1)].
           Subcase B: If A[i_k] = x, then the test "A[i] = x" succeeds and
              t_{k+1} = t_k + 1. Since t_k is the number of occurences of
              x in A[f...(i_k - 1)] by the Ind. Hyp. and A[i_k] = x,
              t_{k+1} = t_k + 1 is the number of occurences of x in 
              A[f...i_k] = A[f...(i_{k+1} - 1)].
           In both cases, LI_{k+1} is true.
  Hence, by induction, LI holds after every iteration of the loop.

When the loop terminates, we know that i > l and by LI: i <= l + 1.
This implies that i = l + 1. Also, LI says that t is the number of occurences
of x in A[f...(i - 1)] = A[f...l], which is exactly the postcondition.