Problem 6. [15 points] In a chicken tournament, for every pair of chickens u and u, either u pecks v or v pecks u, but not both. A king is a chicken u such that for every other chicken U, either · pecks u,or u pecks a chicken w and w pecks u Complete the proof of the following theorem. Theorem. If chicken c is pecked, then c is pecked by a king Proof. Let Sc be the set of all chickens pecked by c, and let De be the set of all chickens that peck c. The situation is illustrated below (Hint: Apply the King Chicken Theorem to De. If chicken c is pecked, then the set D is nonempty. Thus, there is a tournament among the chickens in De, which has a king by the King Chicken Theorem. We will show that d is actually a king of the original tournament d pecks every chicken in De(directly or indirectly), since it is a king of D d pecks chicken c directly, since d is in Dc d pecks every chicken in Sc indirectly, since it pecks c and c pecks every chicken inQuiz 1 9 Problem 6. [15 points] In a chicken tournament, for every pair of chickens u and v, either u pecks v or v pecks u, but not both. A king is a chicken u such that for every other chicken v, either • u pecks v, or • u pecks a chicken w and w pecks v. Complete the proof of the following theorem. Theorem. If chicken c is pecked, then c is pecked by a king. Proof. Let Sc be the set of all chickens pecked by c, and let Dc be the set of all chickens that peck c. The situation is illustrated below: S D c c c (Hint: Apply the King Chicken Theorem to Dc.) If chicken c is pecked, then the set Dc is nonempty. Thus, there is a tournament among the chickens in Dc, which has a king by the King Chicken Theorem. We will show that d is actually a king of the original tournament. • d pecks every chicken in Dc (directly or indirectly), since it is a king of Dc. • d pecks chicken c directly, since d is in Dc. • d pecks every chicken in Sc indirectly, since it pecks c and c pecks every chicken in Sc