Consider the following pseudo code, where $x$ and $y$ are positive integers.begin q := 0 r := x while r >= y do begin r := r - y q := q + 1 end endThe post condition that needs to be satisfied after the program terminates is

The algorithm implements integer division of $x$ by $y$ using repeated subtraction, where $q$ represents the quotient and $r$ represents the remainder.

1. Loop Invariant: At the start, $q = 0$ and $r = x$, so the relation:

   $$x = qy + r$$

   holds initially ($x = 0 \cdot y + x$).

   In each iteration:

   $$r_{\text{new}} = r - y, \quad q_{\text{new}} = q + 1$$

   $$q_{\text{new}} y + r_{\text{new}} = (q + 1)y + (r - y) = qy + r = x$$

   Thus, the invariant $x = qy + r$ is preserved across all iterations.

2. Loop Termination Condition: The loop terminates when the condition $r \ge y$ is false, which means:

   $$r < y$$

3. Post-condition: Combining the loop invariant with the negation of the loop guard yields:

   $$\{x = qy + r \wedge r < y\}$$

Hence, option (B) is correct.