Answer from cs61c-cj (Peter Lau 16539384) for Question 4
Since there is no delay, $monitor may have trouble printing correct results.  $monitor prints when one of its signals changes values, but in the second statement, 3 inputs change values without delay.  $monitor may still print the "previous" values since they are so close together.  Another reason would be that this code does not work in real life without delay implemented.