Question 1191579: You are on the island of knights and knaves, where (a) every local is either
a knight or a knave, (b) knights always tell the truth, and (c) knaves always
lie. Using a symbolic technique (truth table or natural deduction), can you
determine who is a knight and who is a knave? (10 pts. for translation, 10 pts.
for truth table/proof and verdict)
You meet three locals: Al, Bob, and Carol. Al says, “I’m a knave only
if Carol is a knight.” Bob says, “I’m a knight if Carol is.” Carol says,
“Neither Al nor Bob is a knight.”
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's how we can analyze the statements using symbolic logic and determine who is a knight and who is a knave:
**1. Translation into Propositional Logic:**
Let A, B, and C represent the statements "Al is a knight," "Bob is a knight," and "Carol is a knight," respectively. Therefore, ¬A, ¬B, and ¬C mean "Al is a knave," "Bob is a knave," and "Carol is a knave," respectively.
* **Al's statement:** ¬A ↔ C (Al is a knave if and only if Carol is a knight)
* **Bob's statement:** B ↔ C (Bob is a knight if and only if Carol is a knight)
* **Carol's statement:** ¬A & ¬B (Neither Al nor Bob is a knight)
**2. Truth Table:**
We need to consider all possible truth values for A, B, and C, and see which combinations satisfy all three statements.
| A | B | C | ¬A | ¬A ↔ C | B ↔ C | ¬A & ¬B |
|---|---|---|---|---|---|---|
| T | T | T | F | F | T | F |
| T | T | F | F | T | F | F |
| T | F | T | F | F | F | F |
| T | F | F | F | T | T | F |
| F | T | T | T | T | T | F |
| F | T | F | T | F | F | F |
| F | F | T | T | T | F | T |
| F | F | F | T | F | T | T |
**3. Analysis and Verdict:**
Looking at the truth table, we're searching for a row where *all three* statements (¬A ↔ C, B ↔ C, and ¬A & ¬B) are true. Only the last row (A=F, B=F, C=F) satisfies all conditions.
* **Row 8:** A is False, B is False, and C is False.
This means:
* Al is a knave (¬A is true).
* Bob is a knave (¬B is true).
* Carol is a knave (¬C is true).
Therefore, Al, Bob, and Carol are all knaves.
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