A conversation between User and Assistant. The user asks a question, and the Assistant solves it. The assistant first thinks about the reasoning process in the mind and then provides the user with the answer.
User: A very special island is inhabited only by knights and knaves. Knights always tell the truth, and knaves always lie. You meet 7 inhabitants: Scarlett, Daniel, Evelyn, Victoria, Abigail, Aiden, and Riley. In Scarlett's words: "Aiden is a knave if and only if Victoria is a knave". "Aiden is a knave and Scarlett is a knight" - Daniel. "Riley is a knave" - Evelyn. "Aiden is a knight and Abigail is a knight," Victoria declared. "Aiden is a knight or Riley is a knave," Abigail mentioned. "Scarlett is a knave," Aiden claimed. Riley said that If Victoria is a knight then Evelyn is a knight. So who is a knight and who is a knave? Show your work in <think> </think> tags. And return the final answer in <answer> </answer> tags, for example <answer> (1) Zoey is a knight
(2) Oliver is a knight
(3)... </answer>.
Assistant: Let me solve this step by step.
<think> 
user
Assume Scarlett is a knight.  No contradiction is found in their claim that Aiden is a knave if and only if Victoria is a knave.
 Aiden cannot be a knight, because this would contradict the claim of their own that Scarlett is a knave.
 Assume Aiden is a knave.  No contradiction is found in their false claim that Scarlett is a knave.
 Victoria cannot be a knight, because this would contradict the claim of Scarlett that Aiden is a knave if and only if Victoria is a knave.
 Assume Victoria is a knave.  No contradiction is found in their false claim that Aiden is a knight and Abigail is a knight.
 Assume Abigail is a knight.  No contradiction is found in their claim that Aiden is a knight or Riley is a knave.
 Riley cannot be a knight, because this would contradict the claim of Abigail that Aiden is a knight or Riley is a knave.
 Riley cannot be a knave, because this would contradict the false claim of their own that If Victoria is a knight then Evelyn is a knight.
 We have exhausted all possibilities for Riley, so let us go back and reconsider Abigail.
 Assume Abigail is a knave.  No contradiction is found in their false claim that Aiden is a knight or Riley is a knave.
 Assume Riley is a knight.  No contradiction is found in their claim that If Victoria is a knight then Evelyn is a knight.
 Evelyn cannot be a knight, because this would contradict the claim of their own that Riley is a knave.
 Assume Evelyn is a knave.  No contradiction is found in their false claim that Riley is a knave.
 Assume Daniel is a knight.  No contradiction is found in their claim that Aiden is a knave and Scarlett is a knight.
 Therefore, (1) Scarlett is a knight
 (2) Daniel is a knight
 (3) Evelyn is a knave
 (4) Victoria is a knave
 (5) Abigail is a knave
 (6) Aiden is a knave
 (7) Riley is a knight
A conversation between User and Assistant. The user asks a question, and the Assistant solves it. The assistant first thinks about the reasoning process in the mind and then provides the user with the answer.
User: A very special island is inhabited only by knights and knaves. Knights always tell the truth, and knaves always lie. You meet 7 inhabitants: Scarlett, Daniel, Evelyn, Victoria, Abigail, Aiden, and Riley. In Scarlett's words: "Aiden is a knave if and only if Victoria is a knave". "Aiden is a knave and Scarlett is a knight" - Daniel. "Riley is a knave" - Evelyn. "Aiden is a knight and Abigail is a knight," Victoria declared. "Aiden is a knight or Riley is a knave," Abigail mentioned. "Scarlett is a knave," Aiden claimed. Riley said that If Victoria is a knight then Evelyn is a knight. So who is a knight and who is a knave? Show your work in <think> </think> tags. And return the final answer in <answer> </answer> tags, for example <answer> (1) Zoey is a knight
(2) Oliver is a knight
(3)... </answer>.
Assistant: Let me solve this step by step.
<think> 
Assume Scarlett is a knight. No contradiction is found in their claim that Aiden is a knave if and only if Victoria is a knave.
Aiden cannot be a knight, because this would contradict the claim of their own that Scarlett is a knave.
Assume Aiden is a knave. No contradiction is found in their false claim that Scarlett is a knave.
Victoria cannot be a knight, because this would contradict the claim of Scarlett that Aiden is a knave if and only if Victoria is a knave.
Assume Victoria is a knave. No contradiction is found in their false claim that Aiden is a knight and Abigail is a knight.
Assume Abigail is a knight. No contradiction is found in their claim that Aiden is a knight or Riley is a knave.
Riley cannot be a knight, because this would contradict the claim of Abigail that Aiden is a knight or Riley is a knave.
Riley cannot be a knave, because this would contradict the false claim of their own that If Victoria is a knight then Evelyn is a knight.
We have exhausted all possibilities for Riley, so let us go back and reconsider Abigail.
Assume Abigail is a knave. No contradiction is found in their false claim that Aiden is a knight or Riley is a knave.
Assume Riley is a knight. No contradiction is found in their claim that If Victoria is a knight then Evelyn is a knight.
Evelyn cannot be a knight, because this would contradict the claim of their own that Riley is a knave.
Assume Evelyn is a knave. No contradiction is found in their false claim that Riley is a knave.
Assume Daniel is a knight. No contradiction is found in their claim that Aiden is a knave and Scarlett is a knight.
Therefore, (1) Scarlett is a knight
(2) Daniel is a knight
(3) Evelyn is a knave
(4) Victoria is a knave
(5) Abigail is a knave
(6) Aiden is a knave
(7) Riley is a knight </think>
<answer> (1) Scarlett is a knight
(2) Daniel is a knight
(3) Evelyn is a knave
(4) Victoria is a knave
(5) Abigail is a knave
(6) Aiden is a knave
(7) Riley is a knight </answer>
Scarlett Daniel Evelyn Victoria Abigail Aiden Riley
(1) Scarlett is a knight
(2) Daniel is a knight
(3) Evelyn is a knave
(4) Victoria is a knave
(5) Abigail is a knave
(6) Aiden is a knave
(7) Riley is a knight
(('<=>', ('lying', 5), ('lying', 3)), ('and', ('lying', 5), ('telling-truth', 0)), ('lying', 6), ('and', ('telling-truth', 5), ('telling-truth', 4)), ('or', ('telling-truth', 5), ('lying', 6)), ('lying', 0), ('->', ('telling-truth', 3), ('telling-truth', 2)))
rule